Question

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

Short answer

The sum of the series, i.e. $\sum _{n=1}^{\infty }\frac{1}{n{2}^n}=0.693$,

Step-by-step solution

Given Information

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

Definition of Sum of a Series

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

Find the sum of the series

Use the Maclaurin series of ( 1 - x ),

$\mathrm{In}(1-x)=\sum _{n=1}^{\infty }\frac{-{\left(x\right)}^n}{n}$

Further, put $x=\frac{1}{2}$and simplify.

$$\begin{gathered} -In(1-x)=\sum _{n=1}^{\infty }\frac{1}{n{2}^n} \\ \sum _{n=1}^{\infty }\frac{1}{n{2}^n}=-In\left(\frac{1}{2}\right) \\ =0.693 \end{gathered}$$

Hence, the sum of the series, i.e. $\sum _{n=1}^{\infty }\frac{1}{n{2}^n}=0.693$,