Question

Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}$.

Short answer

The sum of the series, i.e. $\sum _{n-1}^{\infty }\frac{2^n}{n!}=e^2-1$,

Step-by-step solution

Given Information

The given series, i.e. $\sum _{n-1}^{\infty }\frac{2^n}{n!}$,

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 $e^x$as:

$$\begin{gathered} e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+.... \\ {e}^x=\sum _{n-0}^{\infty }\frac{x^n}{n!} \\ =1+\sum _{n-0}^{\infty }\frac{x^n}{n!} \end{gathered}$$

Further, put $x=2$and simplify, and we get,

$$\begin{gathered} e^2=1+\sum _{n-1}^{\infty }\frac{2^n}{n!} \\ {e}^2-1=\sum _{n-1}^{\infty }\frac{2^n}{n!} \\ \sum _{n-1}^{\infty }\frac{2^n}{n!}=e^2-1 \end{gathered}$$

Hence, the sum of the series, i.e. $\sum _{n-1}^{\infty }\frac{2^n}{n!}=e^2-1$,