Question

Find whether the series whose $n$th term $\frac{(n+1)x^n}{n^2}$ is convergent or divergent.

Answer 1

Let $u_n=\frac{(n+1)x^n}{n^2}$

Here,

$\frac{u_n}{u_{n-1}}=\frac{(n+1)x^n}{n^2}÷\frac{n{x}^{n-1}}{{(n-1)}^2}$

$=\frac{(n+1){(n-1)}^2}{n^3}.x$
$\therefore \underset{n\to \infty }{lim}\frac{u_n}{u_{n-1}}$

$=\underset{n\to \infty }{lim}\frac{(n+1){(n-1)}^2}{n^3}.x$

$=\underset{n\to \infty }{lim}\frac{n^3(1+\frac{1}{n}){(1-\frac{1}{n})}^2}{n^3}.x$

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

$=x$
Hence if $x<1$ the series is convergent;
If $x>1$ the series is divergent
If $x=1$, then $\underset{n\to \infty }{lim}\frac{u_n}{u_{n-1}}=1$.