# Science:Math Exam Resources/Courses/MATH103/April 2017/Question 04/Solution 1

Let

${\displaystyle \sum _{n=1}^{\infty }{\frac {(n-1)^{2}(x^{2}-1)^{n}}{3^{n}n}}=\sum _{n=1}^{\infty }a_{n}}$

Use the Ratio Test. This requires us to calculate ${\displaystyle L=\lim _{n\to \infty }|{\frac {a_{n+1}}{a_{n}}}|}$. Doing so, we obtain,

${\displaystyle L=\lim _{n\to \infty }|{\dfrac {{\frac {n^{2}}{n+1}}({\frac {x^{2}-1}{3}})^{n+1}}{{\frac {(n-1)^{2}}{n}}({\frac {x^{2}-1}{3}})^{n}}}|=\lim _{n\to \infty }{\frac {n^{2}}{(n-1)^{2}}}{\frac {n}{n+1}}|{\frac {x^{2}-1}{3}}|=|{\frac {x^{2}-1}{3}}|}$

One can check that

${\displaystyle |{\frac {x^{2}-1}{3}}|<1{\text{ if }}|x|<2}$

${\displaystyle |{\frac {x^{2}-1}{3}}|=1{\text{ if }}|x|=2}$

${\displaystyle |{\frac {x^{2}-1}{3}}|>1{\text{ if }}|x|>2}$

So by the Ratio Test, the sum converges if ${\displaystyle |x|<2}$ and the sum diverges if ${\displaystyle |x|>2}$.

When ${\displaystyle |x|=2}$ (i.e., ${\displaystyle x=2}$ or ${\displaystyle x=-2}$), the sum becomes

${\displaystyle \sum _{n=1}^{\infty }{\frac {(n-1)^{2}}{n}}}$.

Since

${\displaystyle \lim _{n\to \infty }|a_{n}|=\lim _{n\to \infty }{\frac {(n-1)^{2}}{n}}=\lim _{n\to \infty }{\frac {n-1}{n}}(n-1)=\infty }$,

by the Term Test for divergence, the series diverges when ${\displaystyle |x|=2}$. Thus, the answer is as follows.

${\displaystyle \color {blue}{\text{The series converges when }}-2