Question

Find the sum of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} \pi^{2 n+1}}{3^{2 n+1}(2 n+1) !} $$

Short answer

The sum of the series is \( \frac{\pi}{3 + \pi^2} \).

Step-by-step solution

Identify the first term and the common ratio

The series is given by \( \sum_{n=0}^{\infty} \frac{(-1)^{n} \pi^{2 n+1}}{3^{2 n+1}(2 n+1) !} \). To identify the first term, substitute \( n = 0 \) into the series to get \( a = \frac{\pi}{3} \). The common ratio \( r \) is \( \frac{-\pi^2}{9} \).

Apply the formula for the sum of an infinite geometric series

The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by the formula \( S = \frac{a}{1 - r} \).

Substitute and simplify

Substituting \( a = \frac{\pi}{3} \) and \( r = \frac{-\pi^2}{9} \) into the formula, we get \( S = \frac{\frac{\pi}{3}}{1 - \frac{-\pi^2}{9}} = \frac{\pi}{3 + \pi^2} \).