Chapter 12.5, Problem 40E

To determine

To express the given series using sigma notation.

Answer to Problem 40E

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

Explanation of Solution

Given information:

Given series is,

  $\frac{3}{9\cdot 2}+\frac{8}{27\cdot 6}+\frac{15}{81\cdot 24}+\dots$

Here we can write $\frac{3}{9\cdot 2}=\frac{2^2-1}{3^2\left(2!\right)},\frac{8}{27\cdot 6}=\frac{3^2-1}{3^3\left(3!\right)},\frac{15}{81\cdot 24}=\frac{4^2-1}{3^4\left(4!\right)}$

Given series can be written in form $\left(\frac{n^2-1}{3^n\left(n!\right)}\right)$

Where $n$ can take values from 2to $\infty$ .

Hence series using sigma notation is,

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