Proving that if $\sum a_n$ converges that $\sum a_n^2$ converges too (if $\sum a_n >0$)

Question

We assume that $a_n >0$ for all $n\geq 0$. This is my proof so far:

If $\sum a_n$ is convergent then the limit of the sequence $\sum a_n$ as $n$ goes to infinity is $0$ and so for any $\epsilon>0$ there's an corresponding $N$ such that:

$0<|a_n-0|< \epsilon$ if $n>N$

If $\epsilon=1$, we can say $0<|a_n^2|<|a_n|<\epsilon$ and then by comparison $\sum a_n^2$ is convergent too.

I'm just stuck on the part where epsilon would be bigger than 1? How could I include that in my proof? Thank you in advance!

Answer 1

If you just assume that $\sum a_n > 0$, then the result is not true. For example a counterexample is given by $$\sum_{n\geq 1} \frac{(-1)^{n+1}}{\sqrt{n}}$$

However, if the hypothesis is that $a_n > 0$, then the fact that $\sum a_n$ converges implies that $a_n$ tends to $0$, so in particular, $$a_n^2=o(a_n)$$

so by comparison, $\sum a_n^2$ also converges.

Answer 2

Assuming that $a_n\ge0$, there is an $N$ such that

$$\forall n>N\implies 0\le a_n\le1$$ which implies $$0\le a_n^2\le a_n\le1$$ and by squeezing the sum of squares converges.