Question

Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point.

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

Short answer

The sum of the series, i.e.$\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\mathrm{\pi }}{2}\right)}^{2\mathrm{n}}=0$,

Step-by-step solution

Given Information

The given series, i.e.$\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\mathrm{\pi }}{2}\right)}^{2\mathrm{n}}$,

Definition of Sum of a Series

The total of all the terms in a sequence can be written as the sum of a series.

Find the sum of the series

Use the Maclaurin series of cos x,

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

Further, put $x=\frac{\mathrm{\pi }}{2}$and simplify as:

$$\begin{gathered} \cos \left(\frac{\begin{array}{c}\mathrm{\pi }\end{array}}{2}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\begin{array}{c}\mathrm{\pi }\end{array}}{2}\right)}^{2n} \\ \sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\begin{array}{c}\mathrm{\pi }\end{array}}{2}\right)}^{2n}=\cos \left(\frac{\begin{array}{c}\mathrm{\pi }\end{array}}{2}\right) \\ =0 \end{gathered}$$

Hence, the sum of the series, i.e.$\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\begin{array}{c}\mathrm{\pi }\end{array}}{2}\right)}^{2n}=0$,