Convergence of $\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)$

Question

Show that the series $$\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)$$ converges .

I was thinking about applying the alternating series test. Letting $c_i$ be the general term, I need to show that $|c_1|\ge |c_2|\ge \dots$; the series is alternating, and $c_i\to 0$.

I think monotonicity holds because $e^x, 1/x$ are monotonically increasing on $(0, +\infty)$ and $\sin x$ is monotonically increasing on $(0,1]$.

It is alternating because $r_i=\sin(1/i)\in(0,1]$ so $e^{r_i}\in(1,e]$ and $e^{-r_i}\in[1/e,1)$. So $c_{2i}$ are positive and $c_{2i-1}$ are negative.

Is the above correct? How can I show that $c_i\to 0$?

Answer 1

hint

Taylor expansion gives $$e^{(-1)^i\sin(1/i)}-1=(-1)^i\sin(1/i)+\frac{\sin^2(1/i)}{2}\Bigl(1+\epsilon(i)\Bigr)=u_i+v_i$$

it is a sum of an alternating convergent series $\sum u_i$ and a series whose general term $v_i$ is equivalent to $\frac{1}{2i^2}$ which converges.

their sum is then convergent.