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If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum {a_n}^2$ always convergent? Either prove it or give a counter example.

Im trying in this way, Suppose $a_n \in [0,1] \ \forall\ n.\ $ Then ${a_n}^2\leq a_n\ \forall\ n.$ Therefore by comparison test $\sum {a_n}^2$ converges.

So If $a_n$ has certain restrictions then the result is true. what about the general case?

How to proceed further? Hints will be greatly appreciated.

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marked as duplicate by Git Gud, Winther, Jonas Meyer real-analysis May 14 '15 at 16:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Hint: If $\sum a_n$ converges, then $a_n\to0$. $\endgroup$ – sranthrop May 14 '15 at 15:32
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    $\begingroup$ @sranthrop It's a funny thing, we posted at the same time and it says EXACTLY the same!. $\endgroup$ – Daniel May 14 '15 at 15:33
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    $\begingroup$ $\sum a_n$ convergent implies that $a_n\rightarrow 0$, then you always have $a_n\in [0,1]$ for $n$ large. $\endgroup$ – Juan Pablo Contreras May 14 '15 at 15:33
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    $\begingroup$ Are people still answering this question? $\endgroup$ – Git Gud May 14 '15 at 16:29
  • $\begingroup$ Your reasoning forgets the cases when $a_n$ is negative. Then $a_n^2 > a_n$, even as $a_n$ approaches 0. $\endgroup$ – NovaDenizen May 14 '15 at 16:47
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If $\sum_{n=1}^\infty a_n$ is convergent, then $\lim_{n\to\infty}a_n=0$, hence from somewhere upward we have $0<a_n<1$, now use comparison test...

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Hint: Since, $a_i>0\forall i\ge 1$, $$\sum_{n=1}^N a_i^2\le \left(\sum_{i=1}^N a_i\right)^2\ \forall N\ge 1$$

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    $\begingroup$ I like this argument best. It requires no new work. Since the original series converges, it's sum exists as a real number, so the square does also. The result follows naturally. $\endgroup$ – Alfred Yerger May 14 '15 at 16:12
  • $\begingroup$ counterexampe: Let $a_1 = 10$ and $a_2 = -10$. $\endgroup$ – NovaDenizen May 14 '15 at 16:38
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    $\begingroup$ $a_i > 0$, so the counterexample does not apply. $\endgroup$ – TokenToucan May 14 '15 at 16:46
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HINT:

If $\sum a_n$ converges then $a_n\to 0$

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  • $\begingroup$ Indeed, it is! :) $\endgroup$ – sranthrop May 14 '15 at 17:25
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Looks like there has been some significant editing but to answer the question as given, as I interpret it, does $\sum a_n$ convergent imply $\sum a_n^2$ convergent for $a_n$ not necessarily non-negative?

The answer is no. Take $a_{2n+1} = 1/\sqrt{n}$ and $a_{2n} = -1/\sqrt{n}$. Then the partial sums will be $s_{2n} = 0$ and $s_{2n+1} = 1/\sqrt{n}$. This means $s_n \to 0$ and the series converges.

However, $a_{2n}^2 = 1/n$ and $a_{2n+1}^2 = 1/n$, so $s_{2n} = 2 \sum_{k=1}^n 1/k$ in this case. This is the harmonic series and it diverges.

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You can find some $n_0\in\Bbb N$ such that $0<a_n<1$ for every $n\ge n_0$.

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You may also use the limit comparison test after you get $a_n\to 0$ by the n-th term test.

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