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$sum a_n$ converges iff $sum fraca_n1+a_n$ converges.
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This is a modification of a problem in Rudin.
Let $(a_n)$ be a sequence of positive numbers (that is $a_n geq 0)$. Then $sum a_n$ converges iff $sum fraca_n1+a_n$ converges.
My attempt:
$Rightarrow$ $$fraca_n1+a_n leq a_n$$ and this follows from comparison test.
$Leftarrow$
Since $$a_n = fraca_n1+a_n (1+a_n) = fraca_n1+a_n + fraca_n^21+a_n$$
it suffices to show that $sum fraca_n^21+a_n$ converges. For this, it suffices to show that $(a_n)$ is bounded, because then the result follows from the comparison test. Indeed, let $M$ be an upperbound. Then
$$fraca_n^21+a_n leq fracMa_n1+a_n$$
We will prove that $a_n to 0$, and this will prove the boundedness.
Let $epsilon > 0$. Choose $N$ such that $fraca_n1+a_n < fracepsilon1+ epsilon$ for $n geq N$, which is possible since $fraca_n1+a_n to 0$ since the series converges.
Then, $n geq N$ implies that $a_n < epsilon$ and the result follows.
Is this correct? Is there an easier way?
real-analysis sequences-and-series proof-verification
asked Jul 25 at 21:12
Math_QED
6,35331344
add a comment |Â
up vote
5
down vote
favorite
This is a modification of a problem in Rudin.
Let $(a_n)$ be a sequence of positive numbers (that is $a_n geq 0)$. Then $sum a_n$ converges iff $sum fraca_n1+a_n$ converges.
My attempt:
$Rightarrow$ $$fraca_n1+a_n leq a_n$$ and this follows from comparison test.
$Leftarrow$
Since $$a_n = fraca_n1+a_n (1+a_n) = fraca_n1+a_n + fraca_n^21+a_n$$
it suffices to show that $sum fraca_n^21+a_n$ converges. For this, it suffices to show that $(a_n)$ is bounded, because then the result follows from the comparison test. Indeed, let $M$ be an upperbound. Then
$$fraca_n^21+a_n leq fracMa_n1+a_n$$
We will prove that $a_n to 0$, and this will prove the boundedness.
Let $epsilon > 0$. Choose $N$ such that $fraca_n1+a_n < fracepsilon1+ epsilon$ for $n geq N$, which is possible since $fraca_n1+a_n to 0$ since the series converges.
Then, $n geq N$ implies that $a_n < epsilon$ and the result follows.
Is this correct? Is there an easier way?
real-analysis sequences-and-series proof-verification
asked Jul 25 at 21:12
Math_QED
6,35331344
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This is a modification of a problem in Rudin.
Let $(a_n)$ be a sequence of positive numbers (that is $a_n geq 0)$. Then $sum a_n$ converges iff $sum fraca_n1+a_n$ converges.
My attempt:
$Rightarrow$ $$fraca_n1+a_n leq a_n$$ and this follows from comparison test.
$Leftarrow$
Since $$a_n = fraca_n1+a_n (1+a_n) = fraca_n1+a_n + fraca_n^21+a_n$$
it suffices to show that $sum fraca_n^21+a_n$ converges. For this, it suffices to show that $(a_n)$ is bounded, because then the result follows from the comparison test. Indeed, let $M$ be an upperbound. Then
$$fraca_n^21+a_n leq fracMa_n1+a_n$$
We will prove that $a_n to 0$, and this will prove the boundedness.
Let $epsilon > 0$. Choose $N$ such that $fraca_n1+a_n < fracepsilon1+ epsilon$ for $n geq N$, which is possible since $fraca_n1+a_n to 0$ since the series converges.
Then, $n geq N$ implies that $a_n < epsilon$ and the result follows.
Is this correct? Is there an easier way?
real-analysis sequences-and-series proof-verification
asked Jul 25 at 21:12
Math_QED
6,35331344
This is a modification of a problem in Rudin.
Let $(a_n)$ be a sequence of positive numbers (that is $a_n geq 0)$. Then $sum a_n$ converges iff $sum fraca_n1+a_n$ converges.
My attempt:
$Rightarrow$ $$fraca_n1+a_n leq a_n$$ and this follows from comparison test.
$Leftarrow$
Since $$a_n = fraca_n1+a_n (1+a_n) = fraca_n1+a_n + fraca_n^21+a_n$$
it suffices to show that $sum fraca_n^21+a_n$ converges. For this, it suffices to show that $(a_n)$ is bounded, because then the result follows from the comparison test. Indeed, let $M$ be an upperbound. Then
$$fraca_n^21+a_n leq fracMa_n1+a_n$$
We will prove that $a_n to 0$, and this will prove the boundedness.
Let $epsilon > 0$. Choose $N$ such that $fraca_n1+a_n < fracepsilon1+ epsilon$ for $n geq N$, which is possible since $fraca_n1+a_n to 0$ since the series converges.
Then, $n geq N$ implies that $a_n < epsilon$ and the result follows.
Is this correct? Is there an easier way?
real-analysis sequences-and-series proof-verification
asked Jul 25 at 21:12
Math_QED
6,35331344
edited Jul 25 at 21:44
asked Jul 25 at 21:12
Math_QED
6,35331344
asked Jul 25 at 21:12
Math_QED
6,35331344
asked Jul 25 at 21:12
Math_QED
6,35331344
6,35331344
add a comment |Â
add a comment |Â
3 Answers
3
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up vote
3
down vote
accepted
Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n to 0$ eventually $a_n<1$ and then $$fraca_n^21+a_n<fraca_n1+a_n$$
As an easier way, assuming that $a_n to 0$ (otherwise both don't converge), we can use limit comparison test and since
$$fraca_nleft(fraca_n1+a_nright)=1+a_n to 1$$
we conclude that $sum a_n$ converges $iff sum fraca_n1+a_n$ converges.
answered Jul 25 at 21:16
gimusi
65k73583
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
 |Â
show 11 more comments
up vote
2
down vote
Claim. If $sum_n=1^infty fraca_n1+a_n$ converges then $a_nto 0$.
[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].
Proof of the claim. If $sum_n=1^infty fraca_n1+a_n$ converges, then $sum_n=1^infty left(frac1+a_n1+a_n-frac11+a_nright)$ converges, so $sum_n=1^infty left(1-frac11+a_nright)$ converges. By the $n$-th term test, the sequence $1-frac11+a_n$ must converge to $0$. Thus, $frac11+a_nto 1$ as $ntoinfty$. From here it is easy to see that $1+a_n to 1$ and so $a_nto 0$ as $ntoinfty$.
answered Jul 25 at 22:17
Prism
5,22121773
add a comment |Â
up vote
1
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Assume $sum_n=1^infty fraca_n1+a_n$ converges.
We claim that $a_n to 0$. Let $0 < varepsilon < 1$ and pick $n_0 inmathbbN$ such that $n ge n_0 implies a_n < fracvarepsilon1-varepsilon$. For such $n$ we have
$$a_n < fracvarepsilon1-varepsilon iff a_n(1-varepsilon) < varepsilon iff a_n < varepsilon (1+a_n) iff fraca_n1+a_n < varepsilon$$
Now pick $M > 0$ such that $1+a_n le M, forall ninmathbbN$.
We have
$$sum_n=1^infty a_n le sum_n=1^infty fracMa_n1+a_n$$
which converges so $sum_n=1^infty a_n$ also converges by the comparison test.
answered Jul 25 at 22:00
mechanodroid
22.2k52041
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n to 0$ eventually $a_n<1$ and then $$fraca_n^21+a_n<fraca_n1+a_n$$
As an easier way, assuming that $a_n to 0$ (otherwise both don't converge), we can use limit comparison test and since
$$fraca_nleft(fraca_n1+a_nright)=1+a_n to 1$$
we conclude that $sum a_n$ converges $iff sum fraca_n1+a_n$ converges.
answered Jul 25 at 21:16
gimusi
65k73583
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
 |Â
show 11 more comments
up vote
3
down vote
accepted
Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n to 0$ eventually $a_n<1$ and then $$fraca_n^21+a_n<fraca_n1+a_n$$
As an easier way, assuming that $a_n to 0$ (otherwise both don't converge), we can use limit comparison test and since
$$fraca_nleft(fraca_n1+a_nright)=1+a_n to 1$$
we conclude that $sum a_n$ converges $iff sum fraca_n1+a_n$ converges.
answered Jul 25 at 21:16
gimusi
65k73583
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
 |Â
show 11 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n to 0$ eventually $a_n<1$ and then $$fraca_n^21+a_n<fraca_n1+a_n$$
As an easier way, assuming that $a_n to 0$ (otherwise both don't converge), we can use limit comparison test and since
$$fraca_nleft(fraca_n1+a_nright)=1+a_n to 1$$
we conclude that $sum a_n$ converges $iff sum fraca_n1+a_n$ converges.
answered Jul 25 at 21:16
gimusi
65k73583
Your proof seems fine, maybe for the second implication, we can simply note that, since $a_n to 0$ eventually $a_n<1$ and then $$fraca_n^21+a_n<fraca_n1+a_n$$
As an easier way, assuming that $a_n to 0$ (otherwise both don't converge), we can use limit comparison test and since
$$fraca_nleft(fraca_n1+a_nright)=1+a_n to 1$$
we conclude that $sum a_n$ converges $iff sum fraca_n1+a_n$ converges.
answered Jul 25 at 21:16
gimusi
65k73583
edited Jul 25 at 21:28
answered Jul 25 at 21:16
gimusi
65k73583
answered Jul 25 at 21:16
gimusi
65k73583
answered Jul 25 at 21:16
gimusi
65k73583
65k73583
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
 |Â
show 11 more comments
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
It is not given that $a_n to 0$
– Math_QED
Jul 25 at 21:17
3
3
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
@Math_QED If $a_n not to 0$ both do not converge.
– gimusi
Jul 25 at 21:18
1
1
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
Yes, that might be easier. But I still don't see how $sum a_n/(1+a_n) < infty$ implies that $a_n to 0$ (Is the proof I provided in my attempt correct?). In your response, you said "assuming $a_n to 0$", but this is what has to be proven!
– Math_QED
Jul 25 at 21:32
1
1
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
(1) How exactly does it follow? Is it because $a_n/(1+a_n) to 0 implies a_n to 0$? (2) If yes, is my proof of this correct (see my attempt)?
– Math_QED
Jul 25 at 21:35
5
5
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
If $b_n = fraca_n1+a_n$ then $a_n = fracb_n1-b_n$; so $b_n to 0$ implies $a_n to 0$.
– Daniel Schepler
Jul 25 at 21:57
 |Â
show 11 more comments
up vote
2
down vote
Claim. If $sum_n=1^infty fraca_n1+a_n$ converges then $a_nto 0$.
[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].
Proof of the claim. If $sum_n=1^infty fraca_n1+a_n$ converges, then $sum_n=1^infty left(frac1+a_n1+a_n-frac11+a_nright)$ converges, so $sum_n=1^infty left(1-frac11+a_nright)$ converges. By the $n$-th term test, the sequence $1-frac11+a_n$ must converge to $0$. Thus, $frac11+a_nto 1$ as $ntoinfty$. From here it is easy to see that $1+a_n to 1$ and so $a_nto 0$ as $ntoinfty$.
answered Jul 25 at 22:17
Prism
5,22121773
add a comment |Â
up vote
2
down vote
Claim. If $sum_n=1^infty fraca_n1+a_n$ converges then $a_nto 0$.
[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].
Proof of the claim. If $sum_n=1^infty fraca_n1+a_n$ converges, then $sum_n=1^infty left(frac1+a_n1+a_n-frac11+a_nright)$ converges, so $sum_n=1^infty left(1-frac11+a_nright)$ converges. By the $n$-th term test, the sequence $1-frac11+a_n$ must converge to $0$. Thus, $frac11+a_nto 1$ as $ntoinfty$. From here it is easy to see that $1+a_n to 1$ and so $a_nto 0$ as $ntoinfty$.
answered Jul 25 at 22:17
Prism
5,22121773
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Claim. If $sum_n=1^infty fraca_n1+a_n$ converges then $a_nto 0$.
[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].
Proof of the claim. If $sum_n=1^infty fraca_n1+a_n$ converges, then $sum_n=1^infty left(frac1+a_n1+a_n-frac11+a_nright)$ converges, so $sum_n=1^infty left(1-frac11+a_nright)$ converges. By the $n$-th term test, the sequence $1-frac11+a_n$ must converge to $0$. Thus, $frac11+a_nto 1$ as $ntoinfty$. From here it is easy to see that $1+a_n to 1$ and so $a_nto 0$ as $ntoinfty$.
answered Jul 25 at 22:17
Prism
5,22121773
Claim. If $sum_n=1^infty fraca_n1+a_n$ converges then $a_nto 0$.
[If I understood correctly, this was the only subtle point. The rest of your proof was perfectly clear].
Proof of the claim. If $sum_n=1^infty fraca_n1+a_n$ converges, then $sum_n=1^infty left(frac1+a_n1+a_n-frac11+a_nright)$ converges, so $sum_n=1^infty left(1-frac11+a_nright)$ converges. By the $n$-th term test, the sequence $1-frac11+a_n$ must converge to $0$. Thus, $frac11+a_nto 1$ as $ntoinfty$. From here it is easy to see that $1+a_n to 1$ and so $a_nto 0$ as $ntoinfty$.
answered Jul 25 at 22:17
Prism
5,22121773
answered Jul 25 at 22:17
Prism
5,22121773
answered Jul 25 at 22:17
Prism
5,22121773
answered Jul 25 at 22:17
Prism
5,22121773
5,22121773
add a comment |Â
add a comment |Â
up vote
1
down vote
Assume $sum_n=1^infty fraca_n1+a_n$ converges.
We claim that $a_n to 0$. Let $0 < varepsilon < 1$ and pick $n_0 inmathbbN$ such that $n ge n_0 implies a_n < fracvarepsilon1-varepsilon$. For such $n$ we have
$$a_n < fracvarepsilon1-varepsilon iff a_n(1-varepsilon) < varepsilon iff a_n < varepsilon (1+a_n) iff fraca_n1+a_n < varepsilon$$
Now pick $M > 0$ such that $1+a_n le M, forall ninmathbbN$.
We have
$$sum_n=1^infty a_n le sum_n=1^infty fracMa_n1+a_n$$
which converges so $sum_n=1^infty a_n$ also converges by the comparison test.
answered Jul 25 at 22:00
mechanodroid
22.2k52041
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
add a comment |Â
up vote
1
down vote
Assume $sum_n=1^infty fraca_n1+a_n$ converges.
We claim that $a_n to 0$. Let $0 < varepsilon < 1$ and pick $n_0 inmathbbN$ such that $n ge n_0 implies a_n < fracvarepsilon1-varepsilon$. For such $n$ we have
$$a_n < fracvarepsilon1-varepsilon iff a_n(1-varepsilon) < varepsilon iff a_n < varepsilon (1+a_n) iff fraca_n1+a_n < varepsilon$$
Now pick $M > 0$ such that $1+a_n le M, forall ninmathbbN$.
We have
$$sum_n=1^infty a_n le sum_n=1^infty fracMa_n1+a_n$$
which converges so $sum_n=1^infty a_n$ also converges by the comparison test.
answered Jul 25 at 22:00
mechanodroid
22.2k52041
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume $sum_n=1^infty fraca_n1+a_n$ converges.
We claim that $a_n to 0$. Let $0 < varepsilon < 1$ and pick $n_0 inmathbbN$ such that $n ge n_0 implies a_n < fracvarepsilon1-varepsilon$. For such $n$ we have
$$a_n < fracvarepsilon1-varepsilon iff a_n(1-varepsilon) < varepsilon iff a_n < varepsilon (1+a_n) iff fraca_n1+a_n < varepsilon$$
Now pick $M > 0$ such that $1+a_n le M, forall ninmathbbN$.
We have
$$sum_n=1^infty a_n le sum_n=1^infty fracMa_n1+a_n$$
which converges so $sum_n=1^infty a_n$ also converges by the comparison test.
answered Jul 25 at 22:00
mechanodroid
22.2k52041
Assume $sum_n=1^infty fraca_n1+a_n$ converges.
We claim that $a_n to 0$. Let $0 < varepsilon < 1$ and pick $n_0 inmathbbN$ such that $n ge n_0 implies a_n < fracvarepsilon1-varepsilon$. For such $n$ we have
$$a_n < fracvarepsilon1-varepsilon iff a_n(1-varepsilon) < varepsilon iff a_n < varepsilon (1+a_n) iff fraca_n1+a_n < varepsilon$$
Now pick $M > 0$ such that $1+a_n le M, forall ninmathbbN$.
We have
$$sum_n=1^infty a_n le sum_n=1^infty fracMa_n1+a_n$$
which converges so $sum_n=1^infty a_n$ also converges by the comparison test.
answered Jul 25 at 22:00
mechanodroid
22.2k52041
edited Jul 25 at 22:10
answered Jul 25 at 22:00
mechanodroid
22.2k52041
answered Jul 25 at 22:00
mechanodroid
22.2k52041
answered Jul 25 at 22:00
mechanodroid
22.2k52041
22.2k52041
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
add a comment |Â
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
Thanks. I think there is an issue in your proof if $epsilon geq 1$.
– Math_QED
Jul 25 at 22:05
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
@Math_QED True, just assume $0 < varepsilon < 1$ so that $fracvarepsilon1-varepsilon > 0$.
– mechanodroid
Jul 25 at 22:11
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