# If There Exists a Sequence $a_n$ where it's Limit is zero, is it possible for $(a_n)^2$ to go to zero slower?

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?

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, $$a_k^2\leq a_k$$.

• Your last inequality should not be strict, I think? – Servaes Nov 1 '18 at 17:15
• I do not understand what you mean, but if $a_k=0$ then the last inequality is false. – Servaes Nov 1 '18 at 17:23
• 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. – Servaes Nov 1 '18 at 17:34

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