Question

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

Answer 1

Let $S=\displaystyle\sum_{r=1}^{\infty} \frac{r+2}{2^{r+1} r \cdot(r+1)}=\displaystyle\sum_{r=1}^{\infty} \frac{2(r+1)-r}{2^{r+1} \cdot r \cdot(r+1)}=\displaystyle\sum_{r=1}^{\infty} \frac{1}{2^{r+1}}\left(\frac{2}{r}-\frac{1}{r+1}\right)=$
$\displaystyle\sum_{r=1}^{\infty}\left(\frac{1}{2^r \cdot r}-\frac{1}{2^{r+1}(r+1)}\right)$
$=\underset{n \rightarrow \infty}{\text{Lim}}\left[\left(\frac{1}{2^1 \cdot 1}-\frac{1}{2^2 \cdot 2}\right)+\left(\frac{1}{2^2 \cdot 2}-\frac{1}{2^3 \cdot 3}\right)\right]+\left(\frac{1}{2^3 \cdot 3}-\frac{1}{2^4 \cdot 4}\right)+\ldots+\left(\frac{1}{2^n \cdot n}-\frac{1}{2^{n+1} \cdot(n+1)}\right)$
$=\underset{n \rightarrow \infty}{\text{Lim}} \left(\frac{1}{2}-\frac{1}{2^{n+1} 9 n+1}\right)$
$\therefore S =\frac{1}{2} $ Hence $S ^{-1}=2$