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I am trying to prove that if $a_n$ >= 0 and $\sum a_n$ converges. Show that $\sum a_n^2$ converges. I'm attempting to use the comparison test by setting it up like, 0 <= $\sum a_n$ <= $\sum a_n^2$. I can do this by showing that $ a_n^2$ goes to zero faster than $a_n$, thus all the points in $a_n^2$ <= $a_n$. If that is true then $\sum a_n^2$ <= $\sum a_n$. Thus, by comparison test the $\sum a_n^2$ is convergent.

I feel like there is a counter example to my argument. Although I haven't been able to come up with one. Is there an example where a sequence converges to zero faster than it's square?

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Hint: Since $a_n\geq0$ and $\sum a_n$ converges, we know that for some $N$, we must have $0\leq a_k<1$ for all $k>N$. Thus, multiplying all of the terms in the inequality by $a_k$, we get $0\leq a_k^2\leq a_k<1$. In particular, $a_k^2\leq a_k$.

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  • $\begingroup$ Your last inequality should not be strict, I think? $\endgroup$ – Servaes Nov 1 '18 at 17:15
  • $\begingroup$ I do not understand what you mean, but if $a_k=0$ then the last inequality is false. $\endgroup$ – Servaes Nov 1 '18 at 17:23
  • $\begingroup$ I don't think the sequence in question is assumed to be decreasing; one term being $0$ does not imply the tail being $0$. Anyway +1 for a clear and simple answer. $\endgroup$ – Servaes Nov 1 '18 at 17:34
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An attempt:

$a_n \ge 0. $

Given $ \sum a_n$ converges, this implies

$\lim_{ n \rightarrow \infty} a_n =0.$

'Goes slower':

Assume there is a $n_0 \in \mathbb{Z^+}$ s.t. for

$n \ge n_0$:

$a_n \le a_n^2$, then

$1 \le a_n$ , a contradiction

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