Question

Find whether the following series are convergent or divergent:
$1+\frac{2x}{|\underset{–}{2}}+\frac{3^2.x^2}{|\underset{–}{3}}+\frac{4^3.x^3}{|\underset{–}{4}}+\frac{5^4.x^4}{|\underset{–}{5}}+....$

Answer 1

To find whether the following series are convergent or divergent

$1+\frac{2x}{2!}+\frac{3^2.x^2}{3!}+\frac{4^3.x^3}{4!}+\frac{5^4.x^4}{5!}+.......$

Here,

$a_n=\frac{x^{n-1}.n^{n-1}}{n!}$

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

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

On applying limit, we get

$\underset{n\to \infty }{lim}\frac{a_n}{a_{n+1}}=\frac{1}{ex}$

If $x>\frac{1}{e}$, then the series is convergent

If $x<\frac{1}{e}$, then the series is divergent

If $x=\frac{1}{e}$, then $\underset{n\to \infty }{lim}\frac{a_n}{a_{n+1}}=\frac{e}{{(1+\frac{1}{n})}^{n-1}}=1$

And $\underset{n\to \infty }{lim}{a}_n=0$

Thus, the given series is convergent for $x=\frac{1}{e}$.