Chapter 8.7, Problem 60E

To determine

To Find: The sum of the series

Answer to Problem 60E

Sum of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}=\frac{\sqrt{3}}{2}$

Explanation of Solution

Given information:

Series

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}$

Formula Used:

  $\cos x={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n}}{(2n)!}}$

Calculation:

The given series is ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}$

Rewrite the series as follows

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\left(\frac{\pi }{6}\right)}^{2n}}{(2n)!}}$

On comparing this series to $\cos x={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n}}{(2n)!}}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}=\cos \frac{\pi }{6}$

So, sum of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{6^{2n}(2n)!}}=\frac{\sqrt{3}}{2}$