Question

Question: 59-64 Find the sum of the series.

61. \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{\pi ^{2n + 1}}}}{{{4^{2n + 1}}(2n + 1)!}}} \)

Short answer

The sum of the series is \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{\pi ^{2n + 1}}}}{{{4^{2n + 1}}(2n + 1)!}}} = \frac{{\sqrt 2 }}{2}\)

Step-by-step solution

Step 1: The equation of power series

When getting a problem like this, we want to notice a power series that is similar to the sum we are given. Here, we have \(\left( {2{\rm{ }}n + 1} \right){\rm{ }}!\) in the denominator, which should point us towards the power series for \(\sin x\):

\(\sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{(2n + 1)!}}\)

Step 2: Final proof

Let's transform our series to a similar form:

\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{\pi ^{2n + 1}}}}{{{4^{2n + 1}}(2n + 1)!}}} {\rm{ }} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \times \frac{{{\pi ^{2n + 1}}}}{{{4^{2n + 1}}}} \times \frac{1}{{(2n + 1)!}}\)\( = \sum\limits_{n = 0}^\infty {( - 1)} \frac{{{{(\pi /4)}^{2n + 1}}}}{{(2n + 1)!}}\)\( = \sin (\pi /4)\)\( = \frac{{\sqrt 2 }}{2}\)