Series $ \sum_{n = 0}^\infty \frac{(-1)^n\sin(n)}{n!} $ is absolutely convergent?

Question

I'm having trouble proving the series

$$ \sum_{n = 0}^\infty \frac{(-1)^n\sin(n)}{n!} $$

is absolutely convergent.

My try

I know that the series

$$ \sum_{n = 0}^\infty \frac{\sin(n)}{n!} $$

converges by the comparison test since,

$$|\sin(x)| \le 1,\ \frac{\sin (n)}{n!} < \frac{1}{n!}$$

However, I cannot prove that the series

$$ \sum_{n = 0}^\infty \frac{(-1)^n\sin(n)}{n!} $$

is convergent.

Help would be greatly appreciated.

Answer 1

An absolutely convergent series is convergent. The former is stronger than the latter, so you're done already. This is since

$$ \bigg|\sum_{k=0}^\infty a_k\bigg|\leqslant \sum_{k=0}^\infty \left|a_k\right|. $$

Answer 2

In addition to the previous answers, some known infinite series (for information only):

$$ \sum_{n=0}^\infty (-1)^n \frac{\sin(n x)}{n!}=-\sin(\sin(x))e^{-\cos(x)}$$ $$ \sum_{n=0}^\infty (-1)^n \frac{\cos(n x)}{n!}=\cos(\sin(x))e^{-\cos(x)}$$ $$ \sum_{n=0}^\infty \frac{\sin(n x)}{n!}=\sin(\sin(x))e^{\cos(x)}$$ $$ \sum_{n=0}^\infty \frac{\cos(n x)}{n!}=\cos(\sin(x))e^{\cos(x)}$$

In the particular case $x=1$ : $$ \sum_{n=0}^\infty (-1)^n \frac{\sin(n)}{n!}=-\sin(\sin(1))e^{-\cos(1)}\simeq-0.4343798300611$$