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Convergence radius of a power series
Clash Royale CLAN TAG#URR8PPP
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1
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I came across the following question :
Find the convergence radius for the series $sum_n=1^infty n!x^n!$.
My initial intuition led me to believe it should converge for $xin(-1,1)$.
I would really appreciate if anyone could review my proof and point out if I`m missing anything.
The proof goes as following :
Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $kin mathbbN$, $a_n_m=n_m$.
By Cauchy-Hadamard theorem, $limsup (|a_n|^frac 1 n)=frac 1R$, and by D'Alembert criterion, if the limit $frac a_n_m+1 a_n_m$ exists , it equals to $lim (|a_n_m|^frac 1 n_m)$ , and indeed $lim frac a_n_m+1 a_n_m=frac n_m+1 n_m=1$. Therefore we have $1$ as a partial limit of $|a_n|^frac 1 n$. In addition, for every $nneq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)
From here we can deduce R=1.
Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)
Is my proof correct?
proof-verification convergence power-series
edited Jul 22 at 17:53
Bernard
110k635103
asked Jul 22 at 17:41
Sar
44910
add a comment |Â
up vote
1
down vote
favorite
I came across the following question :
Find the convergence radius for the series $sum_n=1^infty n!x^n!$.
My initial intuition led me to believe it should converge for $xin(-1,1)$.
I would really appreciate if anyone could review my proof and point out if I`m missing anything.
The proof goes as following :
Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $kin mathbbN$, $a_n_m=n_m$.
By Cauchy-Hadamard theorem, $limsup (|a_n|^frac 1 n)=frac 1R$, and by D'Alembert criterion, if the limit $frac a_n_m+1 a_n_m$ exists , it equals to $lim (|a_n_m|^frac 1 n_m)$ , and indeed $lim frac a_n_m+1 a_n_m=frac n_m+1 n_m=1$. Therefore we have $1$ as a partial limit of $|a_n|^frac 1 n$. In addition, for every $nneq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)
From here we can deduce R=1.
Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)
Is my proof correct?
proof-verification convergence power-series
edited Jul 22 at 17:53
Bernard
110k635103
asked Jul 22 at 17:41
Sar
44910
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I came across the following question :
Find the convergence radius for the series $sum_n=1^infty n!x^n!$.
My initial intuition led me to believe it should converge for $xin(-1,1)$.
I would really appreciate if anyone could review my proof and point out if I`m missing anything.
The proof goes as following :
Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $kin mathbbN$, $a_n_m=n_m$.
By Cauchy-Hadamard theorem, $limsup (|a_n|^frac 1 n)=frac 1R$, and by D'Alembert criterion, if the limit $frac a_n_m+1 a_n_m$ exists , it equals to $lim (|a_n_m|^frac 1 n_m)$ , and indeed $lim frac a_n_m+1 a_n_m=frac n_m+1 n_m=1$. Therefore we have $1$ as a partial limit of $|a_n|^frac 1 n$. In addition, for every $nneq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)
From here we can deduce R=1.
Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)
Is my proof correct?
proof-verification convergence power-series
edited Jul 22 at 17:53
Bernard
110k635103
asked Jul 22 at 17:41
Sar
44910
I came across the following question :
Find the convergence radius for the series $sum_n=1^infty n!x^n!$.
My initial intuition led me to believe it should converge for $xin(-1,1)$.
I would really appreciate if anyone could review my proof and point out if I`m missing anything.
The proof goes as following :
Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $kin mathbbN$, $a_n_m=n_m$.
By Cauchy-Hadamard theorem, $limsup (|a_n|^frac 1 n)=frac 1R$, and by D'Alembert criterion, if the limit $frac a_n_m+1 a_n_m$ exists , it equals to $lim (|a_n_m|^frac 1 n_m)$ , and indeed $lim frac a_n_m+1 a_n_m=frac n_m+1 n_m=1$. Therefore we have $1$ as a partial limit of $|a_n|^frac 1 n$. In addition, for every $nneq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)
From here we can deduce R=1.
Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)
Is my proof correct?
proof-verification convergence power-series
edited Jul 22 at 17:53
Bernard
110k635103
asked Jul 22 at 17:41
Sar
44910
edited Jul 22 at 17:53
Bernard
110k635103
edited Jul 22 at 17:53
Bernard
110k635103
edited Jul 22 at 17:53
Bernard
110k635103
110k635103
asked Jul 22 at 17:41
Sar
44910
asked Jul 22 at 17:41
Sar
44910
asked Jul 22 at 17:41
Sar
44910
44910
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that $$sum_n=1^M n!x^n!le sum_n=1^M! n|x|^n$$
So what can we say about the limit case?
answered Jul 22 at 17:58
Holo
4,2312629
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
add a comment |Â
up vote
-3
down vote
It certainly does NOT converge for $|x| geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x gt 0$) by $sum mx^m.$
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $$sum_n=1^M n!x^n!le sum_n=1^M! n|x|^n$$
So what can we say about the limit case?
answered Jul 22 at 17:58
Holo
4,2312629
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
add a comment |Â
up vote
1
down vote
accepted
Note that $$sum_n=1^M n!x^n!le sum_n=1^M! n|x|^n$$
So what can we say about the limit case?
answered Jul 22 at 17:58
Holo
4,2312629
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $$sum_n=1^M n!x^n!le sum_n=1^M! n|x|^n$$
So what can we say about the limit case?
answered Jul 22 at 17:58
Holo
4,2312629
Note that $$sum_n=1^M n!x^n!le sum_n=1^M! n|x|^n$$
So what can we say about the limit case?
answered Jul 22 at 17:58
Holo
4,2312629
answered Jul 22 at 17:58
Holo
4,2312629
answered Jul 22 at 17:58
Holo
4,2312629
answered Jul 22 at 17:58
Holo
4,2312629
4,2312629
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
add a comment |Â
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
The weak inequality holds in the limit case, and the second limit is the derivative of a geometric sum, which converges for $xin(-1,1)$, but would that mean that the series we're interested does NOT converge for $|x|geq 1$ ?(Or do we need to address this case seperately ?) In addition, I`m not sure why this inequality holds true, can you elaborate?
– Sar
Jul 22 at 18:10
1
1
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
@Sar This does only show that the series converges for $xin(-1,1)$, to show that otherwise the series diverge just notice that for $|x|ge 1$ we have $sum m!x^m!ge sum m!$ which you showed is diverge. The inequality holds because take the set $A=n!mid nle M$ and the set $B=nmid nle M!$, notice that $Asubseteq B$, so we have $sum_n=1^M n!x^n!=sum_nin Anx^n$, with that we can notice that $sum_n=1^M! n|x|^n=sum_nin B n|x|^n=sum_nin An|x|^n+mboxpositive things$, like you said in the post we have...
– Holo
Jul 22 at 18:18
1
1
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
... $sum_nin An|x|^n=sum_nin Anx^n$ because $n!$ is even for all $nge2$
– Holo
Jul 22 at 18:18
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
Thank you very much ! Indeed this is a much simpler and more elegant solution.
– Sar
Jul 22 at 18:20
add a comment |Â
up vote
-3
down vote
It certainly does NOT converge for $|x| geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x gt 0$) by $sum mx^m.$
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
add a comment |Â
up vote
-3
down vote
It certainly does NOT converge for $|x| geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x gt 0$) by $sum mx^m.$
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
add a comment |Â
up vote
-3
down vote
up vote
-3
down vote
It certainly does NOT converge for $|x| geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x gt 0$) by $sum mx^m.$
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
It certainly does NOT converge for $|x| geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x gt 0$) by $sum mx^m.$
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
edited Jul 23 at 17:51
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
answered Jul 22 at 17:45
Aaron Meyerowitz
1,320611
1,320611
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
add a comment |Â
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
1
1
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
It coverage for $|x|ge1$? Really? How come, $sum m!x^m!ge sum m!$ for $|x|ge1$
– Holo
Jul 22 at 17:54
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
Sorry, left out a word. My point was that , while the ratio and root tests will certainly give convergence, easiest is to say that it is absolutely convergent on $(-1,1)$ by comparison to $sum nx^n.$
– Aaron Meyerowitz
Jul 23 at 17:55
add a comment |Â
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