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Find the Sum of the Series $$\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n}}{6^{2n}(2n)!}$$

Alright, so I think I may have gotten this problem correct but I'm a little hesitant, so If you could check my work/find where I went wrong that would be wonderful.

(1) It's a $cos x$ series, so I would simplify the series: $$\frac{\pi}{6}\sum_{n=0}^{\infty}\frac{(-1)^n\pi^n}{2n!}$$ (2) Which then I replace the sum of the series by the $cosx$ value: $$\frac{\pi}{6}\cos{\pi}$$ (3) Then I evaluate: $$\frac{\pi}{6}*(-1) = -\frac{\pi}{6}$$

Is this right?

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The easiest way is to construct the series from the original function explicitly $$ \cos u = \sum_{k=0}^\infty \frac{(-1)^n u^{2n}}{(2n)!} $$ In your case, you need to make $u^{2n}$ look like $\pi^{2n}/6^{2n} = (\pi/6)^{2n}$ so we set $u = \pi/6$ and the LHS becomes $$ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}. $$

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  • $\begingroup$ Ah, I see, thanks for your input! $\endgroup$ – Christian Martinez Apr 15 at 15:18

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