Question

Find whether the following series is convergent or divergent:
$\frac{3}{5}{x}^2+\frac{8}{10}{x}^3+\frac{15}{17}{x}^4+...\frac{n^2-1}{n^2+1}{x}^n+...$ $(n>1)$

Answer 1

To find whether the following series is convergent or divergent:

$\frac{3}{5}{x}^2+\frac{8}{10}{x}^3+\frac{15}{17}{x}^4+.....+\frac{n^2-1}{n^2+1}{x}^n+....$ $(n>1)$

Here, $a_n=\frac{n^2-1}{n^2+1}{x}^n$ and $a_{n-1}=\frac{(n-1{)}^2-1}{(n-1{)}^2+1}{x}^{n-1}$

Also,

$\frac{a_n}{a_{n-1}}=\frac{\frac{n^2-1}{n^2+1}{x}^n}{\frac{(n-1{)}^2-1}{(n-1{)}^2+1}{x}^{n-1}}$

$\Rightarrow \underset{n\to \infty }{lim}\frac{a_n}{a_{n-1}}=\underset{n\to \infty }{lim}\frac{\frac{n^2-1}{n^2+1}{x}^n}{\frac{(n-1{)}^2-1}{(n-1{)}^2+1}{x}^{n-1}}$

On applying limit, we have

$\underset{n\to \infty }{lim}\frac{a_n}{a_{n-1}}=x$

So, if $x>1$, then the series is divergent

and if $x<1$, then the series is convergent

If $x=1$, then $\underset{n\to \infty }{lim}{a}_n=\underset{n\to \infty }{lim}\frac{n^2-1}{n^2+1}=1$

Thus, the series is divergent for $x=1$.