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$ displaystyle lim_x to 0+ sum_n=0^infty frac(-1)^nn!^x = frac12$
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up vote
22
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Prove that $$ lim_x to 0+ sum_n=0^infty frac(-1)^nn!^x =
frac12. $$
We know that $$ sum_n=0^infty frac(-1)^nn!^x$$ converges for any $x>0$. So I try to evaluate the limit as $x$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$.
I know that $$sum_n=0^infty frac(-1)^nn! = frac1e.$$
Does it help to solve this problem?
calculus limits summation
edited Jul 30 at 5:18
Alon Amit
10.2k3765
asked Jul 30 at 3:44
Star Chou
403114
add a comment |Â
up vote
22
down vote
favorite
Prove that $$ lim_x to 0+ sum_n=0^infty frac(-1)^nn!^x =
frac12. $$
We know that $$ sum_n=0^infty frac(-1)^nn!^x$$ converges for any $x>0$. So I try to evaluate the limit as $x$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$.
I know that $$sum_n=0^infty frac(-1)^nn! = frac1e.$$
Does it help to solve this problem?
calculus limits summation
edited Jul 30 at 5:18
Alon Amit
10.2k3765
asked Jul 30 at 3:44
Star Chou
403114
Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
1
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
1
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07
add a comment |Â
up vote
22
down vote
favorite
up vote
22
down vote
favorite
Prove that $$ lim_x to 0+ sum_n=0^infty frac(-1)^nn!^x =
frac12. $$
We know that $$ sum_n=0^infty frac(-1)^nn!^x$$ converges for any $x>0$. So I try to evaluate the limit as $x$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$.
I know that $$sum_n=0^infty frac(-1)^nn! = frac1e.$$
Does it help to solve this problem?
calculus limits summation
edited Jul 30 at 5:18
Alon Amit
10.2k3765
asked Jul 30 at 3:44
Star Chou
403114
Prove that $$ lim_x to 0+ sum_n=0^infty frac(-1)^nn!^x =
frac12. $$
We know that $$ sum_n=0^infty frac(-1)^nn!^x$$ converges for any $x>0$. So I try to evaluate the limit as $x$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$.
I know that $$sum_n=0^infty frac(-1)^nn! = frac1e.$$
Does it help to solve this problem?
calculus limits summation
edited Jul 30 at 5:18
Alon Amit
10.2k3765
asked Jul 30 at 3:44
Star Chou
403114
edited Jul 30 at 5:18
Alon Amit
10.2k3765
edited Jul 30 at 5:18
Alon Amit
10.2k3765
edited Jul 30 at 5:18
Alon Amit
10.2k3765
10.2k3765
asked Jul 30 at 3:44
Star Chou
403114
asked Jul 30 at 3:44
Star Chou
403114
asked Jul 30 at 3:44
Star Chou
403114
403114
Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
1
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
1
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07
add a comment |Â
Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
1
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
1
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07
Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
1
1
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
1
1
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
17
down vote
accepted
Our main claim is as follows:
Proposition. Let $(lambda_n)$ be an increasing sequence of positive real numbers. If $(lambda_n)$ satisfies
$$lim_Rtoinfty frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx = alpha tag1 $$
for some $alpha in [0, 1]$, then
$$lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s = alpha tag2 $$
Here, a sequence $(lambda_n)$ is increasing if $lambda_n leq lambda_n+1$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.
Corollary. Let $(lambda_n)$ be an increasing sequence of positive real numbers that satisfy
- $lim_ntoinfty lambda_n = infty$,
- $lim_ntoinfty lambda_n+1/lambda_n = 1$,
- $lambda_2n < lambda_2n+2$ hold for all sufficiently large $n$ and
$$ lim_ntoinfty fraclambda_2n+1 - lambda_2nlambda_2n+2 - lambda_2n = alpha. tag3 $$
Then we have $text(1)$. In particular, the conclusion $text(2)$ of the main claim continues to hold.
Here are some examples:
The choice $lambda_n = log(n+1)$ satisfies the assumptions with $alpha = frac12$. In fact, this reduces to the archetypal example $eta(0) = frac12$.
OP's conjecture is covered by the corollary by choosing $lambda_n = log(n!)$ and noting that $text(3)$ holds with $alpha = frac12$.
If $P$ is a non-constant polynomial such that $lambda_n = P(n)$ is positive, then $(lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $alpha = frac12$.
Proof of Proposition. Write $F(x) = int_0^x left( sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](t) right) , dt$ and note that
beginalign*
sum_n=0^infty (-1)^n e^-lambda_n s
&= sum_n=0^infty int_lambda_2n^lambda_2n+1 s e^-sx , dx
= int_0^infty s e^-sx , dF(x) \
&= int_0^infty s^2 e^-sx F(x) , dx
stackrelu=sx= int_0^infty s F(u/s) e^-u , du.
endalign*
Since $0 leq F(x) leq x$, the integrand of the last integral is dominated by $ue^-u$ uniformly in $s > 0$. Also, by the assupmption $text(1)$, we have $s F(u/s) to alpha u$ as $s to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that
$$ lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s
= int_0^infty alpha u e^-u , du
= alpha, $$
which completes the proof. ////
Proof of Corollary. For each large $R$, pick $N$ such that $lambda_2N leq R leq lambda_2N+2$. Then
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx leq fraclambda_2N+2lambda_2N cdot fracsum_n=0^N (lambda_2n+1 - lambda_2n)sum_n=0^N (lambda_2n+2 - lambda_2n) $$
and this upper bound converges to $alpha$ as $Ntoinfty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx geq fraclambda_2Nlambda_2N+2 cdot fracsum_n=0^N-1 (lambda_2n+1 - lambda_2n)sum_n=0^N-1 (lambda_2n+2 - lambda_2n) $$
proves the desired claim together with the squeezing theorem. ////
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
 |Â
show 2 more comments
up vote
1
down vote
Define $S(x,,y):=sum_nge 0frac(-1)^nn!^xe^-ny$, which converges for any $x>0$ with $yge 0$ and any $y>0$ with $xge 0$. Grandi's series $sum_nge 0(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $pminfty$ either), but it is said to Abel summable to $frac12$ in the sense $lim_yto 0^+S(0,,y)=frac12$, which you can easily prove with geometric series. The proof you're looking for is $$lim_xto 0^+S(x,,0)=lim_xto 0^+lim_yto 0^+S(x,,y)=lim_yto 0^+lim_xto 0^+S(x,,y)=lim_yto 0^+S(0,,y)=frac12.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.
answered Jul 30 at 5:53
J.G.
12.8k11423
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
Our main claim is as follows:
Proposition. Let $(lambda_n)$ be an increasing sequence of positive real numbers. If $(lambda_n)$ satisfies
$$lim_Rtoinfty frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx = alpha tag1 $$
for some $alpha in [0, 1]$, then
$$lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s = alpha tag2 $$
Here, a sequence $(lambda_n)$ is increasing if $lambda_n leq lambda_n+1$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.
Corollary. Let $(lambda_n)$ be an increasing sequence of positive real numbers that satisfy
- $lim_ntoinfty lambda_n = infty$,
- $lim_ntoinfty lambda_n+1/lambda_n = 1$,
- $lambda_2n < lambda_2n+2$ hold for all sufficiently large $n$ and
$$ lim_ntoinfty fraclambda_2n+1 - lambda_2nlambda_2n+2 - lambda_2n = alpha. tag3 $$
Then we have $text(1)$. In particular, the conclusion $text(2)$ of the main claim continues to hold.
Here are some examples:
The choice $lambda_n = log(n+1)$ satisfies the assumptions with $alpha = frac12$. In fact, this reduces to the archetypal example $eta(0) = frac12$.
OP's conjecture is covered by the corollary by choosing $lambda_n = log(n!)$ and noting that $text(3)$ holds with $alpha = frac12$.
If $P$ is a non-constant polynomial such that $lambda_n = P(n)$ is positive, then $(lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $alpha = frac12$.
Proof of Proposition. Write $F(x) = int_0^x left( sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](t) right) , dt$ and note that
beginalign*
sum_n=0^infty (-1)^n e^-lambda_n s
&= sum_n=0^infty int_lambda_2n^lambda_2n+1 s e^-sx , dx
= int_0^infty s e^-sx , dF(x) \
&= int_0^infty s^2 e^-sx F(x) , dx
stackrelu=sx= int_0^infty s F(u/s) e^-u , du.
endalign*
Since $0 leq F(x) leq x$, the integrand of the last integral is dominated by $ue^-u$ uniformly in $s > 0$. Also, by the assupmption $text(1)$, we have $s F(u/s) to alpha u$ as $s to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that
$$ lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s
= int_0^infty alpha u e^-u , du
= alpha, $$
which completes the proof. ////
Proof of Corollary. For each large $R$, pick $N$ such that $lambda_2N leq R leq lambda_2N+2$. Then
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx leq fraclambda_2N+2lambda_2N cdot fracsum_n=0^N (lambda_2n+1 - lambda_2n)sum_n=0^N (lambda_2n+2 - lambda_2n) $$
and this upper bound converges to $alpha$ as $Ntoinfty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx geq fraclambda_2Nlambda_2N+2 cdot fracsum_n=0^N-1 (lambda_2n+1 - lambda_2n)sum_n=0^N-1 (lambda_2n+2 - lambda_2n) $$
proves the desired claim together with the squeezing theorem. ////
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
 |Â
show 2 more comments
up vote
17
down vote
accepted
Our main claim is as follows:
Proposition. Let $(lambda_n)$ be an increasing sequence of positive real numbers. If $(lambda_n)$ satisfies
$$lim_Rtoinfty frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx = alpha tag1 $$
for some $alpha in [0, 1]$, then
$$lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s = alpha tag2 $$
Here, a sequence $(lambda_n)$ is increasing if $lambda_n leq lambda_n+1$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.
Corollary. Let $(lambda_n)$ be an increasing sequence of positive real numbers that satisfy
- $lim_ntoinfty lambda_n = infty$,
- $lim_ntoinfty lambda_n+1/lambda_n = 1$,
- $lambda_2n < lambda_2n+2$ hold for all sufficiently large $n$ and
$$ lim_ntoinfty fraclambda_2n+1 - lambda_2nlambda_2n+2 - lambda_2n = alpha. tag3 $$
Then we have $text(1)$. In particular, the conclusion $text(2)$ of the main claim continues to hold.
Here are some examples:
The choice $lambda_n = log(n+1)$ satisfies the assumptions with $alpha = frac12$. In fact, this reduces to the archetypal example $eta(0) = frac12$.
OP's conjecture is covered by the corollary by choosing $lambda_n = log(n!)$ and noting that $text(3)$ holds with $alpha = frac12$.
If $P$ is a non-constant polynomial such that $lambda_n = P(n)$ is positive, then $(lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $alpha = frac12$.
Proof of Proposition. Write $F(x) = int_0^x left( sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](t) right) , dt$ and note that
beginalign*
sum_n=0^infty (-1)^n e^-lambda_n s
&= sum_n=0^infty int_lambda_2n^lambda_2n+1 s e^-sx , dx
= int_0^infty s e^-sx , dF(x) \
&= int_0^infty s^2 e^-sx F(x) , dx
stackrelu=sx= int_0^infty s F(u/s) e^-u , du.
endalign*
Since $0 leq F(x) leq x$, the integrand of the last integral is dominated by $ue^-u$ uniformly in $s > 0$. Also, by the assupmption $text(1)$, we have $s F(u/s) to alpha u$ as $s to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that
$$ lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s
= int_0^infty alpha u e^-u , du
= alpha, $$
which completes the proof. ////
Proof of Corollary. For each large $R$, pick $N$ such that $lambda_2N leq R leq lambda_2N+2$. Then
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx leq fraclambda_2N+2lambda_2N cdot fracsum_n=0^N (lambda_2n+1 - lambda_2n)sum_n=0^N (lambda_2n+2 - lambda_2n) $$
and this upper bound converges to $alpha$ as $Ntoinfty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx geq fraclambda_2Nlambda_2N+2 cdot fracsum_n=0^N-1 (lambda_2n+1 - lambda_2n)sum_n=0^N-1 (lambda_2n+2 - lambda_2n) $$
proves the desired claim together with the squeezing theorem. ////
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
 |Â
show 2 more comments
up vote
17
down vote
accepted
up vote
17
down vote
accepted
Our main claim is as follows:
Proposition. Let $(lambda_n)$ be an increasing sequence of positive real numbers. If $(lambda_n)$ satisfies
$$lim_Rtoinfty frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx = alpha tag1 $$
for some $alpha in [0, 1]$, then
$$lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s = alpha tag2 $$
Here, a sequence $(lambda_n)$ is increasing if $lambda_n leq lambda_n+1$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.
Corollary. Let $(lambda_n)$ be an increasing sequence of positive real numbers that satisfy
- $lim_ntoinfty lambda_n = infty$,
- $lim_ntoinfty lambda_n+1/lambda_n = 1$,
- $lambda_2n < lambda_2n+2$ hold for all sufficiently large $n$ and
$$ lim_ntoinfty fraclambda_2n+1 - lambda_2nlambda_2n+2 - lambda_2n = alpha. tag3 $$
Then we have $text(1)$. In particular, the conclusion $text(2)$ of the main claim continues to hold.
Here are some examples:
The choice $lambda_n = log(n+1)$ satisfies the assumptions with $alpha = frac12$. In fact, this reduces to the archetypal example $eta(0) = frac12$.
OP's conjecture is covered by the corollary by choosing $lambda_n = log(n!)$ and noting that $text(3)$ holds with $alpha = frac12$.
If $P$ is a non-constant polynomial such that $lambda_n = P(n)$ is positive, then $(lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $alpha = frac12$.
Proof of Proposition. Write $F(x) = int_0^x left( sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](t) right) , dt$ and note that
beginalign*
sum_n=0^infty (-1)^n e^-lambda_n s
&= sum_n=0^infty int_lambda_2n^lambda_2n+1 s e^-sx , dx
= int_0^infty s e^-sx , dF(x) \
&= int_0^infty s^2 e^-sx F(x) , dx
stackrelu=sx= int_0^infty s F(u/s) e^-u , du.
endalign*
Since $0 leq F(x) leq x$, the integrand of the last integral is dominated by $ue^-u$ uniformly in $s > 0$. Also, by the assupmption $text(1)$, we have $s F(u/s) to alpha u$ as $s to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that
$$ lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s
= int_0^infty alpha u e^-u , du
= alpha, $$
which completes the proof. ////
Proof of Corollary. For each large $R$, pick $N$ such that $lambda_2N leq R leq lambda_2N+2$. Then
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx leq fraclambda_2N+2lambda_2N cdot fracsum_n=0^N (lambda_2n+1 - lambda_2n)sum_n=0^N (lambda_2n+2 - lambda_2n) $$
and this upper bound converges to $alpha$ as $Ntoinfty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx geq fraclambda_2Nlambda_2N+2 cdot fracsum_n=0^N-1 (lambda_2n+1 - lambda_2n)sum_n=0^N-1 (lambda_2n+2 - lambda_2n) $$
proves the desired claim together with the squeezing theorem. ////
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
Our main claim is as follows:
Proposition. Let $(lambda_n)$ be an increasing sequence of positive real numbers. If $(lambda_n)$ satisfies
$$lim_Rtoinfty frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx = alpha tag1 $$
for some $alpha in [0, 1]$, then
$$lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s = alpha tag2 $$
Here, a sequence $(lambda_n)$ is increasing if $lambda_n leq lambda_n+1$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.
Corollary. Let $(lambda_n)$ be an increasing sequence of positive real numbers that satisfy
- $lim_ntoinfty lambda_n = infty$,
- $lim_ntoinfty lambda_n+1/lambda_n = 1$,
- $lambda_2n < lambda_2n+2$ hold for all sufficiently large $n$ and
$$ lim_ntoinfty fraclambda_2n+1 - lambda_2nlambda_2n+2 - lambda_2n = alpha. tag3 $$
Then we have $text(1)$. In particular, the conclusion $text(2)$ of the main claim continues to hold.
Here are some examples:
The choice $lambda_n = log(n+1)$ satisfies the assumptions with $alpha = frac12$. In fact, this reduces to the archetypal example $eta(0) = frac12$.
OP's conjecture is covered by the corollary by choosing $lambda_n = log(n!)$ and noting that $text(3)$ holds with $alpha = frac12$.
If $P$ is a non-constant polynomial such that $lambda_n = P(n)$ is positive, then $(lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $alpha = frac12$.
Proof of Proposition. Write $F(x) = int_0^x left( sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](t) right) , dt$ and note that
beginalign*
sum_n=0^infty (-1)^n e^-lambda_n s
&= sum_n=0^infty int_lambda_2n^lambda_2n+1 s e^-sx , dx
= int_0^infty s e^-sx , dF(x) \
&= int_0^infty s^2 e^-sx F(x) , dx
stackrelu=sx= int_0^infty s F(u/s) e^-u , du.
endalign*
Since $0 leq F(x) leq x$, the integrand of the last integral is dominated by $ue^-u$ uniformly in $s > 0$. Also, by the assupmption $text(1)$, we have $s F(u/s) to alpha u$ as $s to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that
$$ lim_sto0^+ sum_n=0^infty (-1)^n e^-lambda_n s
= int_0^infty alpha u e^-u , du
= alpha, $$
which completes the proof. ////
Proof of Corollary. For each large $R$, pick $N$ such that $lambda_2N leq R leq lambda_2N+2$. Then
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx leq fraclambda_2N+2lambda_2N cdot fracsum_n=0^N (lambda_2n+1 - lambda_2n)sum_n=0^N (lambda_2n+2 - lambda_2n) $$
and this upper bound converges to $alpha$ as $Ntoinfty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound
$$ frac1R int_0^R sum_n=0^infty mathbf1_[lambda_2n, lambda_2n+1](x) , dx geq fraclambda_2Nlambda_2N+2 cdot fracsum_n=0^N-1 (lambda_2n+1 - lambda_2n)sum_n=0^N-1 (lambda_2n+2 - lambda_2n) $$
proves the desired claim together with the squeezing theorem. ////
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
edited Jul 30 at 19:17
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
answered Jul 30 at 11:22
Sangchul Lee
85.5k12155253
85.5k12155253
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
 |Â
show 2 more comments
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
1
1
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
My calculation suggests $lim_ntoinftyfraclambda_2n+1-lambda_2nlambda_2n+2-lambda_2n+1=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$?
– J.G.
Jul 30 at 11:40
1
1
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@J.G. You are right, that is one crucial typo :s I fixed it now.
– Sangchul Lee
Jul 30 at 11:45
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
@i707107, Thank you for proofreading and pointing out typos. I fixed them. :)
– Sangchul Lee
Jul 30 at 14:24
1
1
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
By the same method, for any $kgeq 1$, $$lim_xrightarrow 1- sum_n=0^infty(-1)^n x^n^k=frac 12.$$
– i707107
Jul 30 at 14:51
1
1
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
@Saudman97, This is the indicator notation. For a subset $A$ of $mathbbR$, we define $$mathbf1_A(x) = begincases 1, & textif x in A \ 0, & textif x notin A endcases $$
– Sangchul Lee
Aug 1 at 5:56
 |Â
show 2 more comments
up vote
1
down vote
Define $S(x,,y):=sum_nge 0frac(-1)^nn!^xe^-ny$, which converges for any $x>0$ with $yge 0$ and any $y>0$ with $xge 0$. Grandi's series $sum_nge 0(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $pminfty$ either), but it is said to Abel summable to $frac12$ in the sense $lim_yto 0^+S(0,,y)=frac12$, which you can easily prove with geometric series. The proof you're looking for is $$lim_xto 0^+S(x,,0)=lim_xto 0^+lim_yto 0^+S(x,,y)=lim_yto 0^+lim_xto 0^+S(x,,y)=lim_yto 0^+S(0,,y)=frac12.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.
answered Jul 30 at 5:53
J.G.
12.8k11423
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
 |Â
show 1 more comment
up vote
1
down vote
Define $S(x,,y):=sum_nge 0frac(-1)^nn!^xe^-ny$, which converges for any $x>0$ with $yge 0$ and any $y>0$ with $xge 0$. Grandi's series $sum_nge 0(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $pminfty$ either), but it is said to Abel summable to $frac12$ in the sense $lim_yto 0^+S(0,,y)=frac12$, which you can easily prove with geometric series. The proof you're looking for is $$lim_xto 0^+S(x,,0)=lim_xto 0^+lim_yto 0^+S(x,,y)=lim_yto 0^+lim_xto 0^+S(x,,y)=lim_yto 0^+S(0,,y)=frac12.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.
answered Jul 30 at 5:53
J.G.
12.8k11423
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Define $S(x,,y):=sum_nge 0frac(-1)^nn!^xe^-ny$, which converges for any $x>0$ with $yge 0$ and any $y>0$ with $xge 0$. Grandi's series $sum_nge 0(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $pminfty$ either), but it is said to Abel summable to $frac12$ in the sense $lim_yto 0^+S(0,,y)=frac12$, which you can easily prove with geometric series. The proof you're looking for is $$lim_xto 0^+S(x,,0)=lim_xto 0^+lim_yto 0^+S(x,,y)=lim_yto 0^+lim_xto 0^+S(x,,y)=lim_yto 0^+S(0,,y)=frac12.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.
answered Jul 30 at 5:53
J.G.
12.8k11423
Define $S(x,,y):=sum_nge 0frac(-1)^nn!^xe^-ny$, which converges for any $x>0$ with $yge 0$ and any $y>0$ with $xge 0$. Grandi's series $sum_nge 0(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $pminfty$ either), but it is said to Abel summable to $frac12$ in the sense $lim_yto 0^+S(0,,y)=frac12$, which you can easily prove with geometric series. The proof you're looking for is $$lim_xto 0^+S(x,,0)=lim_xto 0^+lim_yto 0^+S(x,,y)=lim_yto 0^+lim_xto 0^+S(x,,y)=lim_yto 0^+S(0,,y)=frac12.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.
answered Jul 30 at 5:53
J.G.
12.8k11423
answered Jul 30 at 5:53
J.G.
12.8k11423
answered Jul 30 at 5:53
J.G.
12.8k11423
answered Jul 30 at 5:53
J.G.
12.8k11423
12.8k11423
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
 |Â
show 1 more comment
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
1
1
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there?
– Alon Amit
Jul 30 at 10:45
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
@AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $eta(0)=1/2$), but Cesà ro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that.
– J.G.
Jul 30 at 10:53
1
1
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more.
– Alon Amit
Jul 30 at 10:56
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $lambda_2n+2-lambda_2napprox 2(lambda_2n+1-lambda_2n)$ is to be expected when we linearise.
– J.G.
Jul 30 at 11:48
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
@J.G. For many other cases, $lambda_2n+2-lambda_2n$ may not asymptotic to $alpha(lambda_2n+1-lambda_2n)$. For example, $lambda_n=exp(n)$.
– i707107
Jul 30 at 14:49
 |Â
show 1 more comment
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Is this homework? Is there any context where the problem came up?
– abiessu
Jul 30 at 3:48
1
We know that $displaystyle sum_n=0^infty frac(-1)^nn!^k$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $displaystyle frac12$, but I couldn't figure out why.
– Star Chou
Jul 30 at 3:54
You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.)
– saulspatz
Jul 30 at 3:57
1
Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$.
– saulspatz
Jul 30 at 4:07