Question

Let $S$ denote sum of the series $\frac{3}{2^3}+\frac{4}{2^4\cdot 3}+\frac{5}{2^6\cdot 3}+\frac{6}{2^7\cdot 5}+\dots \dots \dots \dots \dots \dots \dots \dots$ Compute the value of $S^{-1}$.

Answer 1

Answer: 0002

Let $S=\sum _{r=1}^{\infty }\frac{r+2}{2^{r+1}r\cdot (r+1)}=\sum _{r=1}^{\infty }\frac{2(r+1)-r}{2^{r+1}\cdot r\cdot (r+1)}=\sum _{r=1}^{\infty }\frac{1}{2^{r+1}}\left(\frac{2}{r}-\frac{1}{r+1}\right)=$
$\sum _{r=1}^{\infty }\left(\frac{1}{2^r\cdot r}-\frac{1}{2^{r+1}(r+1)}\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\left[\left(\frac{1}{2^1\cdot 1}-\frac{1}{2^2\cdot 2}\right)+\left(\frac{1}{2^2\cdot 2}-\frac{1}{2^3\cdot 3}\right)\right]+\left(\frac{1}{2^3\cdot 3}-\frac{1}{2^4\cdot 4}\right)+\dots +\left(\frac{1}{2^n\cdot n}-\frac{1}{2^{n+1}\cdot (n+1)}\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{1}{2}-\frac{1}{2^{n+1}9n+1}\right)$
$\therefore S=\frac{1}{2}$ Hence $S^{-1}=2$