Can anyone help me with the following question.
Any idea where to start. My idea was to try using comparison test since $\sqrt{a_{n}} \leq a_{n}$ but it appears that it wouldn't work if $0 \leq a_{n} <1$.
$$\Bigl(\sqrt a_n-\frac{1}{n}\Bigr)^2=a_n-2*\frac{\sqrt a_n}{n}+\frac{1}{n^2}<a_n+\frac{1}{n^2}.$$
Since all terms are greater than $0$, by the comparison theorem $$\sum_{n=1}^{n=\infty} \Bigl(a_n-2*\frac{\sqrt a_n}{n}+\frac{1}{n^2}\Bigr)$$ converges. But we were given that $\sum_{n=1}^{n=\infty} a_n$ converges and we know $\sum_{n=1}^{n=\infty} \frac{1}{n^2}$ converges, so since the sum or difference of convergent series also converges,the series $\sum_{n=1}^{n=\infty} \frac{\sqrt a_n}{n}$ converges.
I agree with Etheory's answer that the Schwarz inequality is the easiest.
There,
$$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n} \leq \sqrt{\sum_{n=1}^\infty a_n \sum_{n=1}^\infty \frac{1}{n^2}}<\infty.$$
$S_n < (s_n(1 + 1/2^2 + ...+ 1/n^2))^{1/2}$ by Cauchy-Schwarz inequality. $s_n$ is a convergent sequence and being the partial sum of the series $a_n$ hence bounded above by $K$, and $1 + 1/2^2 + ...1/n^2 < \pi^2/6$. So $S(n)$ being the partial sum of the current series is bounded above by $3K$, and it is an increasing sequence so must converge and we're done.
It suffices to show that the sequence
$$
s_n=\sum_{k=1}^n \frac{\sqrt{a_k}}{k}, \quad k\in\mathbb N,
$$
is bounded. (Since $\{s_n\}$ is increasing, and assuming boundedness, we obtain that
$\{s_n\}$ converges.)
Using Cauchy-Schwarz we get that $$ s_n^2=\bigg(\sum_{k=1}^n \frac{a_k}{\sqrt{k}}\bigg)^2\le \bigg(\sum_{k=1}^n\frac{1}{k^2}\bigg)\bigg(\sum_{k=1}^n a_k \bigg)\le \frac{\pi^2}{6}\sum_{k=1}^\infty a_k, \tag{1} $$ since $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$. Since the right hand side of $(1)$ is bounded, so is the left hand side, and hence $\{s_n\}$ is bounded and thus convergent.
