# $\sum_{n=1}^\infty \big(\sin(\sin n)\big)^n$

I'm trying to evaluate the following series: $$\sum_{n=1}^\infty \big(\sin(\sin n)\big)^n$$ In this case the terms are not all positive and I don't know if I can your the usual tests. $$\lim_{n\rightarrow \infty} \big(\sin(\sin n)\big)^n=0$$ because $\sin n$ is between $1$ and $-1$ , $\sin(\sin n)$ is between 1 and -1 , and $\big(\sin(\sin n)\big)^n$ even smaller and between 1 and -1 too. Now I don't know what test to use, I tried but with no result.

• What are sen and senn? Jan 15 '18 at 19:10
• Most people will find the notation $$\sum^\infty_{n=1}(\sin(\sin n))^n$$ a bit more understandable, I guess.
– user436658
Jan 15 '18 at 19:10
• Professor.Very much so.Not just a bit:)) Jan 15 '18 at 19:14
• Do you just need to know if it converges to some number $S$ or you need to get an exact value for $S$? Jan 15 '18 at 19:15
• Assume sen stands for sin, you can use the fact $|\sin(\sin n)| < \sin(1)$ for all $n$ to conclude the series converges absolutely. Jan 15 '18 at 19:15

First, remind that $$\sum |a_n|\text{ converges} \quad \implies \quad \sum a_n \text{ converges}.$$
Then use your usual tests to prove the convergence of $\sum |a_n|$.
Moreover, remind that $\sin(x)$ increases when $x\in [-\pi, \pi]\supset [-1,1]$. And is also true that $-1\leq \sin n \leq 1$.
Then $\sin(-1)\leq \sin\big(\sin n)\big) \leq \sin 1,$ and prove and use the fact that $\sin(1)<1$.