# Series Convergence of $\sum\limits_{n=1}^{\infty} \frac{\cos(n\pi/3)}{n!}$ by Ratio Test

I need to determine whether the series $\sum\limits_{n=1}^{\infty} \frac{\cos(n\pi/3)}{n!}$ converges, specifically by using the ratio test. My problem comes from the fact that that the limit of the cosine function doesn't seem to exist, nor does taking its absolute value make things any better. All I have been able to do so far is set the problem up and cancel the factorial term: \begin{align*} \lim_{n \to \infty} \left\lvert \frac{\frac{\cos((n+1)\pi/3)}{(n+1)!}}{\frac{\cos(n\pi/3)}{n!}}\right\rvert = \lim_{n \to \infty} \left\lvert \frac{\frac{\cos((n+1)\pi/3)}{(n+1)n!}}{\frac{\cos(n\pi/3)}{n!}}\right\rvert = \lim_{n \to \infty} \left\lvert \frac{\frac{\cos((n+1)\pi/3)}{(n+1)}}{\cos(n\pi/3)}\right\rvert \end{align*} At this point I'm stuck. Plotting the cosine function doesn't seem to help much: it seems to oscillate without a clear limit. I'm sure there is some kind of comparison test I could run, but the problem I'm working on specifically requires that I use the ratio test. (I do wonder, though, whether I could use the ratio test within a comparison test to meet the requirements for the problem. Unless I'm missing something, it seems that's the only real possibility.)

I'd greatly appreciate any insights and assistance. Thanks in advance.

• To conclude... don't forget that you do have a factor $n+1$ in the denominator. (But caveat: make sure/mention, when you take the ratio, that you never divide by $0$, i.e. that the $\cos$ never cancels) – Clement C. Dec 22 '17 at 9:01

Hint. Note that $\cos(n\pi/3)\in \{-1,1,-1/2,1/2\}$, hence $$\left|\frac{\cos((n+1)\pi/3)}{\cos(n\pi/3)}\right|\leq 2$$
HINT: $$\left|\dfrac{\cos y}{n!}\right|\leq \frac{1}{n!}.$$ and use the ratio test for $$\sum_{n=1}^{\infty}\frac1{n!}.$$
Hint 1: If you must use only the ratio test, then remember that $\cos\left(\frac{n\pi}3\right)$ has a period of $6$, so work $n\equiv k\pmod6$ separately for each $k\in\{0,1,2,3,4,5\}$.
Hint 2: Easier might be to use comparison against the series $\sum\limits_{n=0}^\infty\frac1{n!}$, to which it is easy to apply the ratio test.