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I'm having trouble with this statement:

"If $$\sum_{n=1}^\infty a_n$$ converges, and $a_n>0$ for all $n$, then $$\sum_{n=1}^\infty a_n^2$$ also converges"

I need to find either a proof or a counterexample. If somebody could give an idea, or a hint, it would be great.

Thanks so much!

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If $\sum_{n=1}^{\infty} a_n $ converges, then this means, there is some $N$ such that $\forall n > N, a_n < 1$. Then, since $0 < a_n^2 < a_n, \forall n > N$, we have that $$\sum_{n=1}^{\infty} a_n^2 = \sum_{n=1}^{N} a_n^2 + \sum_{n=N+1}^{\infty} a_n^2 \le C + \sum_{n=N+1}^{\infty} a_n,$$ where $C<\infty$ is a constant, and $C = \sum_{n=1}^{N} a_n^2$. So, we have that $\sum_{n=1}^{\infty} a_n^2$ converges.

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  • $\begingroup$ Thank you so much! I´ll accept the answer as soon as they allow me to do so. $\endgroup$
    – Lessa121
    Feb 14 '14 at 5:42
  • $\begingroup$ I think a better exercise for you to think now is, what can you say about $a_n^i$ for $i > 0$? $\endgroup$
    – IAmNoOne
    Feb 14 '14 at 5:42
  • $\begingroup$ Well, I can apply the same principle, right? They all converge, even $a_n^n$ $\endgroup$
    – Lessa121
    Feb 14 '14 at 5:45
  • $\begingroup$ No problem. But, they don't quite all converge. The bigger $i$ is the faster they converge. (Assuming convergence of the original.) But, what happens when $i$ gets small? Consider the example $\sum_{n=1}^{\infty} \frac{1}{n^2}$. $\endgroup$
    – mlg4080
    Feb 14 '14 at 5:46
  • $\begingroup$ I was thinking of $i$ as an integer, but if it is in the interval [0,1], then not always, I guess. I could look for a counterexample, maybe $\frac 1 {n^2}$, $i=1/2$ $\endgroup$
    – Lessa121
    Feb 14 '14 at 5:48
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Another way: We know that $$\lim a_n=0.$$ Moreover, $$\lim\frac{{a_n}^2}{a_n}=\lim a_n=0$$ too. Therefore, the results follows by Limit Comperison Test.

Note that: If $a_n \ge 0$, then the mlg4080's answer fits but this does not, directly.

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  • $\begingroup$ this is not limit conparison test,for that test $lim(u_n/v_n)=l\neq 0$. $\endgroup$ Dec 27 '19 at 11:21

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