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

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.

• You need to show $\sum\limits_{n=0}^\infty \Bigl|{(-1)^n\sin n\over n!}\Bigr| = \sum\limits_{n=0}^\infty {|\sin n|\over n!}$ converges. Use the Comparison Test as you did. – David Mitra Aug 5 '14 at 10:14
• @DavidMitra but isn't that only one part of it? don't I need to show that the above series still converges. In order for it to absolutely converge, we need the series $a_n$ and $|a_n|$ to converge correct? – Varun Iyer Aug 5 '14 at 10:21
• By definition of "absolutely convergent", all you need to show is what I said in my previous comment. It is incidental to the problem that any absolutely convergent series is convergent. – David Mitra Aug 5 '14 at 10:23
• So because I showed that |an| converges, I don't need to prove an converges? – Varun Iyer Aug 5 '14 at 10:24
• You showed that your series is absolutely convergent and from that follows that the original series is also convergent. Absolute convergence of a series implies "normal" convergence (That is why the question is only asking for absolute convergence). – ChocolateBar Aug 5 '14 at 10:29

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

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

• So because I showed that $|a_n|$ converges, I don't need to prove $a_n$ converges? – Varun Iyer Aug 5 '14 at 10:22
• Exactly (if you ad the sum in your comment). – frog Aug 5 '14 at 10:34

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

• thanks! Could you also show me a proof :D – Varun Iyer Aug 5 '14 at 10:56
• My aim was only to show some beautifull series, not the proof ! They are also similar series with $\cosh(nx)$ and $\sinh(nx)$. They can be derived from $$\sum_{n=0}^\infty \frac{\exp(n x)}{n!}=e^{e^x}$$ I am pleased to let the people interested to find the proof by themself or in the books. – JJacquelin Aug 5 '14 at 12:23