Mixing
If $sum (a_n)^2$ converges then $ sum frac{a_n}{n} $ converges (one solution)
$begingroup$
My solution is detailed, I would like to know if it is correct or not.
As the series converges and each $(a_n)^2$ is positive, then exists $K> 0$ such that : $(a_1)^2+(a_2)^2+ cdots + (a_n)^2 < K, forall n in mathbb{N}$.
Given $n in mathbb{N}$ : $ lvert dfrac{a_n}{n} rvert = dfrac{lvert a_n rvert}{n} leq dfrac{(a_n)^2}{n} leq (a_n)^2$ so :
$dfrac{lvert a_1 rvert}{1} + dfrac{lvert a_2 rvert}{2} + cdots dfrac{lvert a_n rvert}{n} leq (a_1)^2+(a_2)^2 + cdots (a_n)^2<K$
Finally the series $sum dfrac{lvert a_n vert}{n}$ converge, consequently $sum dfrac{a_n}{n} $ converge.
real-analysis sequences-and-series convergence
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
$endgroup$
|
show 1 more comment
$begingroup$
My solution is detailed, I would like to know if it is correct or not.
As the series converges and each $(a_n)^2$ is positive, then exists $K> 0$ such that : $(a_1)^2+(a_2)^2+ cdots + (a_n)^2 < K, forall n in mathbb{N}$.
Given $n in mathbb{N}$ : $ lvert dfrac{a_n}{n} rvert = dfrac{lvert a_n rvert}{n} leq dfrac{(a_n)^2}{n} leq (a_n)^2$ so :
$dfrac{lvert a_1 rvert}{1} + dfrac{lvert a_2 rvert}{2} + cdots dfrac{lvert a_n rvert}{n} leq (a_1)^2+(a_2)^2 + cdots (a_n)^2<K$
Finally the series $sum dfrac{lvert a_n vert}{n}$ converge, consequently $sum dfrac{a_n}{n} $ converge.
real-analysis sequences-and-series convergence
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
$endgroup$
1
$begingroup$
This is a shining example of the use of the Cauchy--Schwarz inequality.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 20:36
$begingroup$
I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
$endgroup$
– User8128
Jan 27 at 20:37
$begingroup$
$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
$endgroup$
– Ixion
Jan 27 at 20:39
$begingroup$
If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
$endgroup$
– Calvin Khor
Jan 27 at 21:08
$begingroup$
@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
$endgroup$
– user
Jan 27 at 21:25
|
show 1 more comment
$begingroup$
My solution is detailed, I would like to know if it is correct or not.
As the series converges and each $(a_n)^2$ is positive, then exists $K> 0$ such that : $(a_1)^2+(a_2)^2+ cdots + (a_n)^2 < K, forall n in mathbb{N}$.
Given $n in mathbb{N}$ : $ lvert dfrac{a_n}{n} rvert = dfrac{lvert a_n rvert}{n} leq dfrac{(a_n)^2}{n} leq (a_n)^2$ so :
$dfrac{lvert a_1 rvert}{1} + dfrac{lvert a_2 rvert}{2} + cdots dfrac{lvert a_n rvert}{n} leq (a_1)^2+(a_2)^2 + cdots (a_n)^2<K$
Finally the series $sum dfrac{lvert a_n vert}{n}$ converge, consequently $sum dfrac{a_n}{n} $ converge.
real-analysis sequences-and-series convergence
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
$endgroup$
My solution is detailed, I would like to know if it is correct or not.
As the series converges and each $(a_n)^2$ is positive, then exists $K> 0$ such that : $(a_1)^2+(a_2)^2+ cdots + (a_n)^2 < K, forall n in mathbb{N}$.
Given $n in mathbb{N}$ : $ lvert dfrac{a_n}{n} rvert = dfrac{lvert a_n rvert}{n} leq dfrac{(a_n)^2}{n} leq (a_n)^2$ so :
$dfrac{lvert a_1 rvert}{1} + dfrac{lvert a_2 rvert}{2} + cdots dfrac{lvert a_n rvert}{n} leq (a_1)^2+(a_2)^2 + cdots (a_n)^2<K$
Finally the series $sum dfrac{lvert a_n vert}{n}$ converge, consequently $sum dfrac{a_n}{n} $ converge.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
edited Jan 27 at 20:44
José Carlos Santos
170k23132238
170k23132238
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
asked Jan 27 at 20:33
Juan Daniel Valdivia FuentesJuan Daniel Valdivia Fuentes
758
758
1
$begingroup$
This is a shining example of the use of the Cauchy--Schwarz inequality.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 20:36
$begingroup$
I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
$endgroup$
– User8128
Jan 27 at 20:37
$begingroup$
$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
$endgroup$
– Ixion
Jan 27 at 20:39
$begingroup$
If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
$endgroup$
– Calvin Khor
Jan 27 at 21:08
$begingroup$
@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
$endgroup$
– user
Jan 27 at 21:25
|
show 1 more comment
1
$begingroup$
This is a shining example of the use of the Cauchy--Schwarz inequality.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 20:36
$begingroup$
I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
$endgroup$
– User8128
Jan 27 at 20:37
$begingroup$
$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
$endgroup$
– Ixion
Jan 27 at 20:39
$begingroup$
If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
$endgroup$
– Calvin Khor
Jan 27 at 21:08
$begingroup$
@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
$endgroup$
– user
Jan 27 at 21:25
1
1
$begingroup$
This is a shining example of the use of the Cauchy--Schwarz inequality.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 20:36
$begingroup$
This is a shining example of the use of the Cauchy--Schwarz inequality.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 20:36
$begingroup$
I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
$endgroup$
– User8128
Jan 27 at 20:37
$begingroup$
I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
$endgroup$
– User8128
Jan 27 at 20:37
$begingroup$
$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
$endgroup$
– Ixion
Jan 27 at 20:39
$begingroup$
$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
$endgroup$
– Ixion
Jan 27 at 20:39
$begingroup$
If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
$endgroup$
– Calvin Khor
Jan 27 at 21:08
$begingroup$
If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
$endgroup$
– Calvin Khor
Jan 27 at 21:08
$begingroup$
@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
$endgroup$
– user
Jan 27 at 21:25
$begingroup$
@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
$endgroup$
– user
Jan 27 at 21:25
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
No, that is not correct. You have no reason to assume that$$(forall ninmathbb{N}):frac{lvert a_nrvert}nleqslant{a_n}^2.$$
The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $displaystylesum_{n=1}^inftyfrac1{n^2}$.
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
add a comment |
$begingroup$
Another way can be by using the AM-GM inequality: For positive $x,y$ we have $frac{x^2+y^2}{2} geq xy $. Now put $x=|a_k|$ and $y=dfrac{1}{k}$. Then, we have
$$ a_k^2 + frac{1}{k^2} geq 2frac{|a_k|}{k} $$
Adding up, we see that
$$ sum_{k=1}^n a_k^2 + sum_{k=1}^n frac{1}{k^2} geq 2 sum frac{ |a_k| }{k} $$
can you finish it?
edited Jan 27 at 21:27
user
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
No, that is not correct. You have no reason to assume that$$(forall ninmathbb{N}):frac{lvert a_nrvert}nleqslant{a_n}^2.$$
The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $displaystylesum_{n=1}^inftyfrac1{n^2}$.
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
add a comment |
$begingroup$
No, that is not correct. You have no reason to assume that$$(forall ninmathbb{N}):frac{lvert a_nrvert}nleqslant{a_n}^2.$$
The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $displaystylesum_{n=1}^inftyfrac1{n^2}$.
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
add a comment |
$begingroup$
No, that is not correct. You have no reason to assume that$$(forall ninmathbb{N}):frac{lvert a_nrvert}nleqslant{a_n}^2.$$
The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $displaystylesum_{n=1}^inftyfrac1{n^2}$.
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
No, that is not correct. You have no reason to assume that$$(forall ninmathbb{N}):frac{lvert a_nrvert}nleqslant{a_n}^2.$$
The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $displaystylesum_{n=1}^inftyfrac1{n^2}$.
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 20:36
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
add a comment |
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
Yes, you are right. Is true but if $lvert a_n rvert > 1$, thanks! :)
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:40
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
$begingroup$
@JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge.
$endgroup$
– user
Jan 27 at 21:31
add a comment |
$begingroup$
Another way can be by using the AM-GM inequality: For positive $x,y$ we have $frac{x^2+y^2}{2} geq xy $. Now put $x=|a_k|$ and $y=dfrac{1}{k}$. Then, we have
$$ a_k^2 + frac{1}{k^2} geq 2frac{|a_k|}{k} $$
Adding up, we see that
$$ sum_{k=1}^n a_k^2 + sum_{k=1}^n frac{1}{k^2} geq 2 sum frac{ |a_k| }{k} $$
can you finish it?
edited Jan 27 at 21:27
user
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
add a comment |
$begingroup$
Another way can be by using the AM-GM inequality: For positive $x,y$ we have $frac{x^2+y^2}{2} geq xy $. Now put $x=|a_k|$ and $y=dfrac{1}{k}$. Then, we have
$$ a_k^2 + frac{1}{k^2} geq 2frac{|a_k|}{k} $$
Adding up, we see that
$$ sum_{k=1}^n a_k^2 + sum_{k=1}^n frac{1}{k^2} geq 2 sum frac{ |a_k| }{k} $$
can you finish it?
edited Jan 27 at 21:27
user
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
add a comment |
$begingroup$
Another way can be by using the AM-GM inequality: For positive $x,y$ we have $frac{x^2+y^2}{2} geq xy $. Now put $x=|a_k|$ and $y=dfrac{1}{k}$. Then, we have
$$ a_k^2 + frac{1}{k^2} geq 2frac{|a_k|}{k} $$
Adding up, we see that
$$ sum_{k=1}^n a_k^2 + sum_{k=1}^n frac{1}{k^2} geq 2 sum frac{ |a_k| }{k} $$
can you finish it?
edited Jan 27 at 21:27
user
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
$endgroup$
Another way can be by using the AM-GM inequality: For positive $x,y$ we have $frac{x^2+y^2}{2} geq xy $. Now put $x=|a_k|$ and $y=dfrac{1}{k}$. Then, we have
$$ a_k^2 + frac{1}{k^2} geq 2frac{|a_k|}{k} $$
Adding up, we see that
$$ sum_{k=1}^n a_k^2 + sum_{k=1}^n frac{1}{k^2} geq 2 sum frac{ |a_k| }{k} $$
can you finish it?
edited Jan 27 at 21:27
user
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
edited Jan 27 at 21:27
user
5,92011031
edited Jan 27 at 21:27
user
5,92011031
edited Jan 27 at 21:27
user
5,92011031
5,92011031
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
answered Jan 27 at 20:40
Jimmy SabaterJimmy Sabater
2,560325
2,560325
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
add a comment |
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
$begingroup$
Yes, the series $sum (a_n)^2 $ and $sum dfrac{1}{n^2}$ converge. So the serie $sum dfrac{lvert a_n rvert}{n}$ converge. But I can not use the fact that the series $sum dfrac{1}{n^2}$, converges because I had been told to find a way without using it.
$endgroup$
– Juan Daniel Valdivia Fuentes
Jan 27 at 20:47
add a comment |
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1
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This is a shining example of the use of the Cauchy--Schwarz inequality.
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– Pedro Tamaroff♦
Jan 27 at 20:36
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I don’t see how you can say $lvert a_nrvert / n le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality.
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– User8128
Jan 27 at 20:37
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$|a_n|le a_n^2iff a_nle -1 vee a_n=0 vee a_nge 1$
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– Ixion
Jan 27 at 20:39
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If $sum a_n^2$ converges, then so does $sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ sum frac{|a_n| }{n} le left (sum |a_n|^3 right )^{1/3} left( sumfrac 1{n^{3/2}} right)^{2/3} <infty $$ without using $sum frac{1}{n^2} < infty$...
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– Calvin Khor
Jan 27 at 21:08
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@CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$.
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– user
Jan 27 at 21:25