Sum of series $\sum_1 \frac{\cos(x)^{n-1}}{n!}$

I'm asked to find when this series converges and the sum of this series:

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

I found out, with the ratio criteria that it converges everywhere but when $x=0+k\pi$ but I have no idea how to find the sum of this series.

• It also converges at $x=0+k\pi$. You can verify directly. – Dunham Feb 20 '18 at 20:43
• \begin{eqnarray*} e^{\cos(x)}=1+ \cos(x) \left( \sum_{n=1}^{\infty} \frac{\cos^{n-1}(x)}{n!} \right). \end{eqnarray*} – Donald Splutterwit Feb 20 '18 at 20:44
• I think you mean everywhere but $x=\pi/2 + k\pi$ – saulspatz Feb 20 '18 at 20:46
• @saulspatz Also there is convergence. In fact, the series converges everywhere on the real line: it is just a particular case of the well known series for the exponential function $\;e^x\;$ . – DonAntonio Feb 20 '18 at 20:51
• @DonAntonio Well, the limits of summation aren't given. I assumed OP was taking the lower limit of summation to be $n=0$ and worrying about where $\cos x = 0.$ Otherwise, the series obviously converges everywhere on $\mathbb R,$ and I couldn't imagine what the problem was. – saulspatz Feb 20 '18 at 20:57

If $x=2k\pi$ the series becomes $$\sum \frac {1}{n!}$$ which converges by ratio test and the sum is $e$
If $x=(2k+1)\pi$, it becomes $\sum \frac {(-1)^{n-1}}{n!}$ which converges by alternate criteria and the sum is $e^{-1}$.
• So it converges everywhere? @saulspatz says that it doesn't converge in $x=\pi/2+k\pi$... – KatherineEnilin Feb 20 '18 at 20:52
• @KatherineEnilin No, that was a misunderstanding. After your clarification, I agree it converges absolutely on $\mathbb R$. – saulspatz Feb 20 '18 at 21:14
$\mid \cos x \mid \le 1,$ so the series converges absolutely everywhere on $\mathbb R$ by comparison with the series $\sum{\frac{1}{n!}}=e.$