Stuck at proving convergence of the series that is dependent on a converging series

Suppose $\sum_{n=1}^{\infty}{a_n}$ converges, and $a_n > 0$. Does $$\sum_{n=1}^{\infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}}$$ converge or diverge?

Attempt: I was able to prove that it diverges, as shown below, but could not find an example.

Claim: $\sum_{n=1}^{\infty}{\dfrac{1}{\sqrt{n}+na_n}}$ diverges. Proof: Since $\sum_{n=1}^{\infty}{a_n}$ converges, there exists a $n\geq n_0$ such that $$0 \leq a_n \leq 1$$ which gives, $$\dfrac{1}{\sqrt{n}+na_n} \geq \dfrac{1}{\sqrt{n}+n}$$ proving the claim. Doing a limit comparison test for$\sum_{n=1}^{\infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}}$ with $\sum_{n=1}^{\infty}{\dfrac{1}{\sqrt{n}+na_n}}$ we get $$\lim_{n\rightarrow \infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}\cdot \dfrac{\sqrt{n}+na_n}{1}} = \sin(\sqrt{a_n}) < \infty$$ and hence the given series diverges. However, I am having trouble finding an example.

• But $\sin \sqrt{a_n} \to 0$, that can easily force convergence (consider $a_n = \frac{1}{n^2}$). To see if/that is always the case is not so easy/obvious. – Daniel Fischer Jul 15 '13 at 15:13
• Right, but the series may fail to converge even if $\lim_{n\rightarrow \infty}{b_n}=0$ where $b_n$ is the series in question. – AAP Jul 15 '13 at 15:19
• It may, but it does not always. I have an example of $a_n$ with finite sum so that the modified series diverges. – Daniel Fischer Jul 15 '13 at 15:21
Hint: It is easy to see it converges for some $(a_n)$. To see it could diverge, you may consider $a_n=\frac{1}{n(\log n)^2}$ for $n\ge 2$.