Question

Sum the following series
$1+\frac{3}{4}+\frac{7}{16}+\frac{15}{64}+\frac{31}{256}+....$ to infinity.

Answer 1

We have

$S=1+\frac{3}{4}+\frac{7}{16}+\frac{15}{64}+\frac{31}{256}\dots \dots \left(1\right)$

Multiplying $S$ by $4$, we get

$4S=4+3+\frac{7}{4}+\frac{15}{16}+\frac{31}{64}\dots \dots \left(2\right)$

$\left(2\right)-\left(1\right)$

$\Rightarrow 3S=6+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots \cdots =6+2=8$

$\Rightarrow S=\frac{8}{3}$