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I need to find the value of the following serie:

$ \sum_{n=0}^\infty \frac{1}{2^n}+\frac{(\pi i)^n}{n!}$

Our professor didn't show us how to do that with complex numbers.
$\frac{1}{2^n}$ gets smaller and smaller as n increases. When I expand $\frac{(\pi i)^n}{n!}$, I get $1+\pi i -\pi^2/2- i \pi^3/6 + \pi^4/24+ i\pi^5/120 - \pi^6/720 - ... $

As expected, the sign changes every two times because of the imaginary number, but I don't really see what to do next.

Thanks for your help.

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    $\begingroup$ $e^z=\sum \frac {z^n}{n!}$ even for complex $z$. $\endgroup$ – lulu Feb 21 '18 at 19:22
  • $\begingroup$ If I distribute the sum, I get 2 + e^(pi*i), because 1/2^n can be rewritten as (1/2)^n which is a geometric sum and converges to 2. Is that correct ? $\endgroup$ – Poujh Feb 21 '18 at 19:32
  • $\begingroup$ Yes, that is correct. $\endgroup$ – lulu Feb 21 '18 at 20:26

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