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$\sum_{n=0}^{\infty} \frac{\cos{(n\pi/3)}}{n!}$ = $\sqrt{e} \cos(\sqrt{3}/2)$

Wolfram alpha gave me this answer for the sum. First I'd like to know if I can use the alternating series test to prove that this sum is convergent (absolutely or conditionally). Then I'd like to know how wolfram came up with this answer for the sum.

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  • $\begingroup$ Brief notes on convergence: (1) Absolute convergence is easy to show directly, (2) The alternating series test does not indicate absolute convergence, (3) Dirichlet's test is a generalization of the alternating series test you can use to show convergence. $\endgroup$ – Jonas Meyer Apr 8 '15 at 3:39
  • $\begingroup$ If you know $\exp (i \theta) $ then it is trivial. Otherwise I think you can use telescoping sum, though I am not sure. $\endgroup$ – aNumosh Apr 8 '15 at 3:40
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The easiest way to get the sum is using Euler's formula $e^{i\theta} = \cos\theta+i\sin\theta$. What you have is the real part of $\sum_{n=0}^\infty \frac{1}{n!} \bigl( e^{i \pi / 3} \bigr)^n$, which is $\exp(e^{i\pi/3})$. Then, $e^{i\pi/3}=\frac{1+i\sqrt{3}}{2}$, so $\exp(e^{i\pi/3})=e^{1/2}\, e^{i \sqrt{3}/2}$, and taking the real part gives the cosine.

What you need to prove, though, doesn't require any of that. Remember that the cosine is bounded between $-1$ and $+1$, and that $n!$ grows very fast. In fact, $\sum_{n=0}^\infty \frac{1}{n!} = e$ is certainly convergent, without any need for alternating properties. The alternating series test is not the best way to go.

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\begin{eqnarray} \sum_{n=0}^\infty\frac{\cos(n\pi/3)}{n!}&=&\Re\sum_{n=0}^\infty\frac{e^{in\pi/3}}{n!}=\Re\exp(e^{i\pi/3})=\Re\exp(\cos\pi/3+i\sin\pi/3)\\ &=&\Re\exp(1/2+i\sqrt{3}/2)=e^{1/2}\cos(\sqrt{3}/2) \end{eqnarray}

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    $\begingroup$ The $\Re$ stands for the 'real part' i.e. the real part of $e^{ix}$ is $\cos(x)$ as it has no imaginary aspect. Similarly, $\Im$ stands for the 'imaginary part', which in the case of $e^{ix}$ is $\sin(x)$. $\endgroup$ – Mattos Apr 8 '15 at 3:52

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