Prove convergence of $\sum_{n=1}^{\infty}\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}$

Question

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.

Answer 1

Hint: $\sin (\frac {\pi n} {n+5})=\sin (\pi -\frac {\pi n} {n+5})=\sin (\frac {5\pi } {n+5})$ and $|\sin (\frac {5\pi } {n+5})| \leq \frac {5\pi } {n+5}$. Now compare the sreies with $\sum \frac 1 {n^{3}}$.

Answer 2

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.