Question

$\frac{1}{n+1}+\frac{1}{2(n+1{)}^2}+\frac{1}{3(n+1{)}^3}+...$ = a $\times$ $\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-...$ Find a

Answer 1

Given
$\frac{1}{n+1}+\frac{1}{2(n+1{)}^2}+\frac{1}{3(n+1{)}^3}+...=a\times \frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-...$ ....(1)
Consider, $\frac{1}{n+1}+\frac{1}{2(n+1{)}^2}+\frac{1}{3(n+1{)}^3}+...$
$=-(-\frac{1}{n+1}-\frac{1}{2(n+1{)}^2}-\frac{1}{3(n+1{)}^3}+...$
$=-{\log }_e(1-\frac{1}{n+1})$
$={\log }_e\left(\frac{n+1}{n}\right)$
$={\log }_e(1+\frac{1}{n})$ ....(2)
Now consider, $\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-...$
$={\log }_e(1+\frac{1}{n})$
So,from equation (1),(2) and (3), we get
${\log }_e(1+\frac{1}{n})=a{\log }_e(1+\frac{1}{n})$
$\Rightarrow a=1$