help with proving convergence or divergence of series I need to prove the convergence or divergence of:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}$$
I've tried:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!} = \lim_{n \to \infty} \biggl|\frac{(-1)^n\cos(\frac{\pi (n+1)}{3})}{(n+1)!} \frac{n!}{(-1)^{n-1}\cos(\frac{\pi n}{3})}\biggl|$$
Which was working until I was left with:
$$\lim_{n->\infty} \frac{\cos(\frac{\pi (n+1)}{3})}{(n+1)\cos(\frac{n\pi}{3})}$$
So I gave up on that route and tried comparison test:
$$a_n = \frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!} < \frac{(-1)^{n-1}}{n!} = b_n$$
Then did the ratio test on $b_n$:
$$L = \lim_{n->\infty} \frac{n!}{(n+1)!} = 0$$
So: $L < 1$ means $b_n$ converges and $a_n < b_n$ so $a_n$ also converges.
I took a shot in the dark with this one. Could someone let me know if I'm on the right track?
Tanks :)
 A: Actually in your argument $\frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}<\frac{(-1)^{n-1}}{n!}$ is not true, since these are alternating terms. However, it is true once you take the absolute values. Then you can show $a_n$ converges absolutely.
More precicesly, set $a_n = \frac{(-1)^{n-1}cos(\frac{\pi n}{3})}{n!}$, so $|a_n| \le \frac{1}{n!}$. By ratio test $\sum_{n=1}^{\infty}\frac{1}{n!}$ converges. Hence by comparison test, $\sum_{n=1}^{\infty} |a_n|$ converges, and thus $\sum_{n=1}^{\infty} a_n$ converges.
A: Comparison test is applied when terms of the series are eventually non-negative. In your case, nth terms of the series  $\frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}$, is not guaranteed to be positive. 
You may proceed as follows: 
$|\frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}|\le \frac{1}{n!}\le \frac{1}{2^{n-1}}$ 
$\sum_{n=1}^\infty \frac{1}{2^{n-1}}$ is convergent. Hence by limit comparison test, $\sum_{n=1}^\infty |\frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}|$ is convergent $\implies$ $\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos(\frac{\pi n}{3})}{n!}$ is convergent (absolutely).
