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Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.

For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent. Not to sure if this is correct, any help would be much appreciated, many thanks.

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  • $\begingroup$ Great question! It is easy to check that this series converges since you know $\ell^2$ is Hilbert. But it is a very hard insight when you think only using the basic real Analysis. $\endgroup$
    – checkmath
    Feb 23, 2012 at 22:14
  • $\begingroup$ Many thanks I really appreciate the help. $\endgroup$
    – user24930
    Feb 24, 2012 at 19:34
  • $\begingroup$ can you explain the transition for which ${a_n}^2 \leq \frac{|a_n|}{n} \Longrightarrow \frac{|a_n|}{n} \leq \frac{1}{n^2} + {a_n}^2$ ? $\endgroup$
    – Noa Even
    Jan 23, 2019 at 14:41
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    $\begingroup$ @Jneven $a_n^2 = |a_n||a_n|$, thus it follows that $a_n^2 \leq \frac{|a_n|}{n} \implies |a_n| \leq \frac{1}{n} \implies \frac{|a_n|}{n} \leq \frac{1}{n^2}$. You can add anything you'd like to the right side as long as it is nonnegative and the inequality will still hold. $\endgroup$
    – cwk
    Jan 28, 2019 at 19:37

6 Answers 6

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What you did is correct; in fact you can show that if $\{a_n\}$ and $\{b_n\}$ are two sequences of real numbers and $\sum_{n\geq 0}a_n^2$ and $\sum_{n\geq 0}b_n^2$ are convergent then the series $\sum_{n=0}^{+\infty}|a_nb_n|$ is convergent, noting that $0\leq |a_nb_n|\leq \max(a_n^2,b_n^2)\leq a_n^2+b_n^2$.

Your particular case is $b_n=\frac 1n$.

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  • $\begingroup$ Many thanks really appreciate it. $\endgroup$
    – user24930
    Feb 24, 2012 at 19:35
  • $\begingroup$ Instead of $\sum {a_n}^2$, Let $a_n$ positive terms. if $\sum_{n=1}^\infty (a_n)^{3/2}<\infty$ then will $\sum_{n=1}^\infty a_n/n$ converges? $\endgroup$
    – user464147
    May 3, 2019 at 16:16
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Another approach is to note that for any positive integer $N,$ we have $\sum_{n=1}^{N} \frac{|a_n|}{n} \leq \sqrt{ \sum_{n= 1}^{N} a_{n}^{2}} \sqrt{ \sum_{n=1}^{N} \frac{1}{n^2}}$, and this is in turn less than $\frac{\pi}{\sqrt{6}}\sqrt{ \sum_{n= 1}^{N} a_{n}^{2} }.$ The first inequality follows by the Cauchy-Schwarz inequality, and the second follows by Euler's formula $\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}.$

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  • $\begingroup$ Many thanks really appreciate it. $\endgroup$
    – user24930
    Feb 24, 2012 at 19:35
  • $\begingroup$ Great answer :) $\endgroup$
    – MathMan
    Jan 27, 2015 at 16:18
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You are right.. you can also simplify things further using that $|ab| \leq {a^2 + b^2\over 2}$ for any $a$ and $b$, so that $|{a_n \over n}| \leq {a_n^2 \over 2} + {1 \over 2n^2}$ and thus your series converges absolutely as you are saying.

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  • $\begingroup$ Many thanks really appreciate it. $\endgroup$
    – user24930
    Feb 24, 2012 at 19:35
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AM-GM:

$b_n := a_n^2+ 1/n^2 \ge 2|a_n|(1/n)$.

$\sum b_n$ converges.

$S_n := 2\sum_{k=1}^{n}|a_n|(1/n)$ is increasing and bounded above, hence converges, which implies

$\sum a_n/n$ converges.

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Note that for every $n\in\mathbb{N}$ we have

$$\left(|a_n|-\frac{1}{n}\right)^2\geq0\Longrightarrow 2\frac{|a_n|}{n}\leq a_n^2+\frac{1}{n^2}.$$ Then we can use the comparison test to assert the convergence of $\sum\dfrac{|a_n|}{n}$. Once again, by the same test, we have that $\sum\dfrac{a_n}{n}$.

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    $\begingroup$ You have added nothing new to already existing answers $\endgroup$
    – Shailesh
    Nov 8, 2019 at 1:56
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Lets split the sum $\sum_{n=1}^\infty a_n^2$ into two sums (we can since it converges absolutely) , one is the sum of all $a_n^2$ where $a_n\leq \frac1n$ and the other of all $a_n^2$ such that $a_n>\frac1n$, now for the first sum we have $\frac{a_n}n\leq \frac1{n^2}$ and for second we have $a_n^2=a_n\cdot a_n>a_n\cdot\frac1n$ so we have that the corresponding two sums of $\frac{a_n} n$ also converge so we can combine them and rearrange (because both converge absolutely) them into $\sum_{n=1}^\infty\frac{a_n}n$

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