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.
Thanks in advance!