$\sum a_n$ converges $\implies\ \sum a_n^2$ converges? [duplicate]

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.

marked as duplicate by Git Gud, Winther, Jonas Meyer real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '15 at 16:56

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

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...

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

• 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. – Alfred Yerger May 14 '15 at 16:12
• counterexampe: Let $a_1 = 10$ and $a_2 = -10$. – NovaDenizen May 14 '15 at 16:38
• $a_i > 0$, so the counterexample does not apply. – TokenToucan May 14 '15 at 16:46

HINT:

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

• Indeed, it is! :) – sranthrop May 14 '15 at 17:25

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.

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

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