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I've been trying to do this problem for so long...

Suppose that $(a_n)_n$ is a sequence with $a_n \geq 0$ for each $n \in \mathbb{N}$, and suppose further that the series $\sum_{n=1}^{\infty}{a_n}$ is convergent. Prove that the series $\sum_{n=1}^{\infty}{{a_n}^2}$ is also convergent.

I've tried to use the ratio test, the comparison test and even epsilon proofs but I'm not getting anywhere. Would be very grateful if somebody could help me.

Thanks,

Henry

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marked as duplicate by user186170, Peter Košinár, saz, MathOverview, JonMark Perry Oct 10 '16 at 19:56

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  • $\begingroup$ Ahh my gosh thanks.. then you use the comparison test to show convergence :) $\endgroup$ – Henry Oct 9 '16 at 21:28
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Hint:

Since $\sum a_n$ converges, $a_n$ approaches 0. Hence there is some $N$ for which $a_n<1$ for all $n>N$. In this regime, $a_n^2\leq a_n$.

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  • $\begingroup$ Actually, $a_{n}^2 \leq a_{n}$ does not follow if $a_{n}$ is negative, only $a_{n}^2 \leq |a_{n}|$ follows. $\endgroup$ – Matija Sreckovic Oct 9 '16 at 21:29
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    $\begingroup$ @Matija Sreckovic: The statement of the problem assumes $a_n\geq0$. $\endgroup$ – symplectomorphic Oct 9 '16 at 21:29
  • $\begingroup$ Oh, very good then. My bad. $\endgroup$ – Matija Sreckovic Oct 9 '16 at 21:30
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Hint: recall (or realise) the following about convergence of series: If there is some $N>0$ such that the partial sums from the $N$th term up to the $n$th term converge as $n\to \infty$ then the series converges.

Now realise that as the series converges, you must have $a_n\to0$ and so there is some term after which all terms satisfy $0\le a_n<1$. What do you know about $x^2$ in relation to $x$ when $0<x<1$?

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  • $\begingroup$ Is there like a theorem for the first paragraph of your comment or is it just so obvious that you don't need to mention it in a proof? $\endgroup$ – Henry Oct 9 '16 at 21:34
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Your statement is actually incorrect, so I guess that's why you were unable to prove it.

Take, for example, the series $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{\sqrt{n}},$$ which converges according to Leibniz's test, but its square, $$\sum_{n=0}^{\infty}\frac{1}{n},$$ diverges.

EDIT: I didn't read the question carefully, and I missed the part $a_{n} \geq 0$. With that assumption, symplectomorphic's solution is what I was going to type. Anyway, if the sequence weren't positive, you'd need the series to be absolutely convergent in order to prove the statement, and my example (although not an answer to your question) is a good counterexample which shows why absolute convergence is required.

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    $\begingroup$ The theorem assumes the terms are nonnegative. $\endgroup$ – symplectomorphic Oct 9 '16 at 21:29

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