Question

Let $S$ denotes the sum of series $\frac{3}{2^3}+\frac{4}{2^4\cdot 3}+\frac{5}{2^6\cdot 3}+\frac{6}{2^7\cdot 5}+\cdots +\infty .$ Then the of $S^{-1}$ is

Answer 1

Let
$S=\sum _{r=1}^{\infty }\frac{r+2}{2^{r+1}\cdot r(r+1)}\Rightarrow S=\sum _{r=1}^{\infty }\frac{2(r+1)-r}{2^{r+1}\cdot r(r+1)}\Rightarrow S=\sum _{r=1}^{\infty }(\frac{1}{2^r\cdot r}-\frac{1}{2^{r+1}\cdot (r+1)})\Rightarrow S=\underset{n\to \infty }{lim}(\frac{1}{2^1\times 1}-\frac{1}{2^2\times 2})+(\frac{1}{2^2\times 2}-\frac{1}{2^3\times 3})+(\frac{1}{2^3\times 3}-\frac{1}{2^4\times 4})+(\frac{1}{2^n\times n}-\frac{1}{2^{n+1}\times (n+1)})\Rightarrow S=\underset{n\to \infty }{lim}(\frac{1}{2}-\frac{1}{2^{n+1}\times (n+1)})\Rightarrow S=\frac{1}{2}\Rightarrow S^{-1}=2$