# Prove $\sum a_n^2$ converges.

Suppose $\left(a_{n}\right)$ is a monotonic sequence of nonnegative real numbers converging to 0. Assume also that the series $\sum\frac{a_{n}}{\sqrt{n}}$ is convergent. Prove that the series $\sum a_{n}^{2}$ is also convergent. Prove that if the monotonicity hypothesis is dropped, then the above conclusion is not true.

If we show that $\sqrt{n}a_{n}\leq B$ for all n for some B , then we have $$\sum a_{n}^{2}=\sum\left(\sqrt{n}a_{n}\right)\left(\frac{a_{n}}{\sqrt{n}}\right)\leq B\sum\left(\frac{a_{n}}{\sqrt{n}}\right)<\infty.$$

So now I want to show that there exists such a $B$. What makes me think that this must hold is that $\lim\frac{a_{n}}{\sqrt{n}}=0$ and hence $a_{n}$ is approching 0 faster than $\sqrt{n}$ is approaching $\infty$ (or equivalently faster than $\frac{1}{\sqrt{n}}$ is approching 0). But how do I show this? Any ideas?

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Since $\sum \frac{a_n}{\sqrt{n}}$ is convergent, there exist $B$ such that for all $n \ge 1$,
$$\sum_{k = 1}^n \frac{a_k}{\sqrt{k}} \le B$$
Now, because $(a_k)$ is decreasing, we have
$$B \ge \sum_{k = 1}^n \frac{a_k}{\sqrt{k}} \ge a_n \sum_{k = 1}^n \frac{1}{\sqrt{k}} \ge a_n \sum_{k = 1}^n \frac{1}{\sqrt{n}} = a_n \sqrt{n}$$
For a counter example in the case $a_n$ is not monotonic, consider for example $a_n = \frac{(-1)^n}{\sqrt{n}}$. EDIT : for a nonnegative example, take $(a_n)$ defined by $a_{2^k} = \frac{1}{\sqrt{k}}$ and $a_n = 0$ if $n$ is not a power of $2$