Question

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

Answer 1

To find whether the following series are convergent or divergent

Given series is

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

Here, $a_n=\frac{n^2}{n!}$

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

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

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

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

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

Finally on applying limit, we have

$L=0$

Since $L<1$

Hence, the given series converges.