# Given that $\sum_{n=1}^\infty a_n$ is convergent prove that $\sum_{n=1}^\infty \left(\frac{1+\sin(a_n)}{2}\right)^n$ also converges.

Show that if $$\sum_{n=1}^\infty a_n$$ converges, then $$\sum_{n=1}^\infty \left(\frac{1+\sin(a_n)}{2}\right)^n$$ converges.

I tried to prove it using the Comparison Test and the Ratio Test but I am not able to come to a conclusive result using those two. I've tried all other convergence tests available to me but I have no luck solving this question. Any guidance would be greatly appreciated.

• Note that $\lim\limits_{n\rightarrow\infty}a_n=0$, so for big $n$ we have $|a_n|<\frac{1}{2}$, so $|\frac{1+\sin a_n}{2}|\leq \frac{3}{4}$. Does it help? Feb 27, 2020 at 18:43

Hint: $$|a_n| \to 0$$, so for sufficiently large $$n$$ you'll have $$|1+\sin(a_n)|/2 < 3/4$$.
The idea will be if $$\sum a_n < \infty$$ then $$a_n \to 0$$, which would mean $$\sin a_n \to 0$$ as well, since for $$x \to 0$$, you have $$\sin x \approx x$$. So for $$n > N$$, each term in your series can be bounded by $$\frac{1 + \sin(a_n)}{2} < \frac{2}{3},$$ or any other number in $$(1/2,1)$$ and now your sum bounds by geometric series.