<s>[INST] Prove convergence of $\sum_{n=1}^{\infty}\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}$ 
I need to prove convergence of the series.
  $$
\sum_{n=1}^{\infty}a_n,\ \ 
a_n=\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}
$$

Well, at first I noticed that $\forall n\in\mathbb{N}\ a_n>0$:
$$
\left.
\begin{aligned}
&0<\frac{\pi n}{n+5}<\pi\Rightarrow\sin^3\left(\frac{\pi n}{n+5}\right)>0\\
&\sqrt{n^2+n}>n\Rightarrow\sqrt{n^2+n}-n>0
\end{aligned}
\right\}\Rightarrow a_n>0
$$
So, then I tried to apply comparison test to this series. However, it was not helpful since no matter how I bounded $\sin^3\left(\frac{\pi n}{n+5}\right)$, I could not find $b_n\geqslant a_n:\sum_{n=1}^{\infty}b_n$ is convergent.
Thus, I would be glad if someone could give me some clue to this problem.
 [/INST] We have that 
$$\sin^3\left(\frac{\pi n}{n+5}\right)=\sin^3\left(\frac{\pi n+\pi 5-\pi 5}{n+5}\right)=\sin^3\left(\pi-\frac{\pi 5}{n+5}\right)=$$
$$=\sin^3\left(\frac{\pi 5}{n+5}\right) \sim \left(\frac{\pi 5}{n+5}\right)^3$$
and
$$\frac{1}{\sqrt{n^2+n}-n}=\frac{1}{\sqrt{n^2+n}-n}\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}=\frac{\sqrt{n^2+n}+n}{n}\sim 2$$
therefore $a_n \sim \frac1{n^3}$ and the given series converges by comparison test.
</s>