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

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?

• You're expression in (1) is wrong - how did you remove $6^{2n}?$ The correct answer is to write out the series for $\cos x$ and then figure out what $x$ is. It is not $\pi.$ – Thomas Andrews Apr 15 at 15:27
• – lab bhattacharjee Apr 15 at 15:33

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 $$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.$$

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