Question

Find whether the following series is convergent or divergent:
$2x+\frac{3x^2}{8}+\frac{4x^3}{27}+...+\frac{(n+1)x^n}{n^3}+...$

Answer 1

Find whether the following series is convergent or divergent

Given series is

$2x+\frac{3x^2}{8}+\frac{4x^3}{27}+.......+\frac{(n+1)x^n}{n^3}+....$

Also, $a_n=\frac{(n+1)x^n}{n^3}$

According to Ratio test

If there exists an $N$ so that $n\ge N$

If $L<1$, then the series converges

If $L>1$, then the series diverges

If $L=1$, then the Ratio test is inconclusive

where $L=\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|$

On applying Ratio test, we get

$\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{lim}\left|\frac{\frac{(n+2)x^{n+1}}{(n+1{)}^3}}{\frac{(n+1)x^n}{n^3}}\right|$

$\Rightarrow L=\underset{n\to \infty }{lim}\left|\frac{n^3(n+2)x^{n+1}}{(n+1{)}^4{x}^n}\right|$

On applying limit, we have

$L=\underset{n\to \infty }{lim}\left|\frac{n^3(n+2)x}{(n+1{)}^4}\right|$

$\Rightarrow L=\left|x\right|$

For convergence

$L<1$

i.e. $\left|x\right|<1$

i.e. $-1\le x\le 1$

Therefore, given series converges for $-1\le x\le 1$ and diverges for $x>1$.