Question

Find whether the following series are convergent or divergent:
$\frac{1^2}{2^2}+\frac{1^2.3^2}{2^2.4^2}x+\frac{1^2.3^2.5^2}{2^2.4^2.6^2}{x}^2+...$

Answer 1

To find whether the following series are Convergent or Divergent

$\frac{1^2}{2^2}+\frac{1^2{.3}^2}{2^2{.4}^2}x+\frac{1^2{.3}^2{.5}^2}{2^2{.4}^2{.6}^2}{x}^2+....$

Here , We have

$a_n=\frac{1^2{.3}^2{.5}^2.....(2n-1{)}^2}{2^2{.4}^2{.6}^2......(2n{)}^2}{x}^{n-1}$

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

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

If $x<1$ , the Series is Convergent

If $x>1$ , the Series is Divergent

If $x=1$ ,

$\underset{n\to \infty }{lim}n(\frac{a_n}{a_{n+1}}-1)=\underset{n\to \infty }{lim}\frac{n(4n+3)}{(2n+1{)}^2}$

Therefore , We have to use Another Test .

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

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

$=\frac{-nlogn}{4n^2}=\frac{-logn}{4n}=0$

Hence , Given Series is Divergent