Chapter 10.4, Problem 64E

To determine

To prove: ${\displaystyle \sum _{n=0}^{\infty }\frac{n^2}{2n}}$

Explanation of Solution

Given information: $\frac{1}{\left(1-x\right)}$

Formula used: ${\displaystyle \sum _{n=0}^{\infty }{x}^n}=\frac{1}{\left(1-x\right)}$

Calculation:

Differentiating the summation formula with respect to $x$ ,

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

Multiplying both sides by $x$ ,

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

By using Quotient Rule on right hand side of the equation differentiate both sides of the equation.

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{n}^2{x}^{n-1}}=\frac{1{\left(1-x\right)}^2-x\times 2\left(1-x\right)\left(-1\right)}{{\left(1-x\right)}^4}\\ {\displaystyle \sum _{n=0}^{\infty }{n}^2{x}^{n-1}}=\frac{1\left(1-x\right)+2x}{{\left(1-x\right)}^3}\\ {\displaystyle \sum _{n=0}^{\infty }{n}^2{x}^{n-1}}=\frac{1+x}{{\left(1-x\right)}^3}\\ {\displaystyle \sum _{n=0}^{\infty }{n}^2{x}^n}=\frac{\left(1+x\right)x}{{\left(1-x\right)}^3}\end{array}$

Rewriting the left hand side of the equation as the summation mention in the question,

  ${\displaystyle \sum _{n=0}^{\infty }{n}^2{\left(\frac{1}{2}\right)}^n}=\frac{\frac{1}{2}\left(1+\frac{1}{2}\right)}{{\left(1-\frac{1}{2}\right)}^3}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n^2}{2^n}}=6$

Hence, proved.