Using the Ratio Test on Trig Functions

I've got to use the Ratio Test to determine whether this series is convergent or divergent:

$$\sum_{n=1}^\infty \frac{cos(n\pi/3)}{n!}$$

Taking the $\lim \limits_{n \to \infty}|\frac{a_n+1}{a_n}|$ gets me as far as $$\lim \limits_{n \to \infty}|\frac{cos((n+1)\pi/3)}{cos(n\pi/3)(n+1)}|$$

I'm not sure how to proceed at this point. Unless I'm wrong, $|cos((n)\pi/3)| \lt 1$ for all $n \ge 1$, which should also go for $|cos((n+1)\pi/3)|$, so I can't just take the $\lim$ as ${n \to \infty}$.

Of course $(n+1) \to \infty$ as $n \to \infty$, meaning that as n becomes large, any possible value of the cosine functions should still result in the ratio becoming infinitesimal (since they're both $\le 1$), but I'm not 100% on that, and even so, I'd rather know what the "official" next steps would be in solving the problem.

• Your teacher is asking you explicitly to use Ratio test? – imranfat Mar 12 '16 at 23:28
• Hint: $(n+1)\pi/3 = n\pi/3 + \pi/3$ – John Joy Mar 12 '16 at 23:29
• use addition formula for cosine – lzralbu Mar 12 '16 at 23:30
• The limit is $0$! You are doing fine! – Mhenni Benghorbal Mar 12 '16 at 23:33

I'd start with the comparison test: $$\Big|\frac{\cos(\frac{n\pi}{3})}{n!}\Big|\leq \frac{1}{n!}$$ and then you can use the ratio test on the series $\sum_n\frac{1}{n!}$
You can use the ratio test. In particular, consider that in the limit $$\lim_{n\rightarrow\infty}\left|\frac{\cos((n+1)\pi/3)}{\cos(n\pi/3)\cdot (n+1)}\right|$$ the only problem we could have would be if the $|\cos(n\pi/3)|$ in the denominator was small (or worse - zero). However, since $\cos$ is periodic, we can find out that this quantity can only ever be $\cos(0)=1$ or $\cos(\pi/3)=\frac{1}2$. Therefore $\cos(n\pi/3)\geq \frac{1}2$ which gives $$\left|\frac{\cos((n+1)\pi/3)}{\cos(n\pi/3)\cdot (n+1)}\right|\leq \left|\frac{\cos((n+1)\pi/3)}{\frac{1}2(n+1)}\right|$$ and then we can just note that $\cos(x)$ is always less than $1$ so this term is less than $\frac{2}{n+1}$ and thus goes to zero. The idea is that you need to establish (upper) bounds on your term rather than working with it directly.
It's worth noting that, if you saw this problem out in the wild and weren't bound to the ratio test, the comparison with $\frac{1}{n!}$ would be a much better idea. For instance, if you had $\sum_{n=1}^{\infty}\frac{\cos(\alpha n)}{n!}$, you'd have quite a lot of trouble using the ratio test, since exactly how the ratio behaves depends on $\alpha$ in some rather deep way, whereas the comparison test obviously gives convergence.