Sum of Weird Fractions

Let $S=\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\dots \infty$

Dividing by $2$ ,

$\frac{S}{2}= \frac{1}{8}+\frac{3}{16}+\frac{5}{32}+\dots \infty$

subtract second equation from the first,

$\Rightarrow \frac{S}{2}= \frac{1}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}\dots \infty$

$\Rightarrow \frac{S}{2}= \frac{1}{4}+\frac{2}{8}(1+\frac{1}{2}+\frac{1}{8}+\frac{1}{16}\dots \infty)$

$\Rightarrow \frac{S}{2}=\frac{1}{4}+\frac{2}{8} \times \frac{1}{1-\frac{1}{2}}$

$\Rightarrow S = \frac{3}{2}$

So the answer is $3-2 = \fbox{1}$

Let $S = \dfrac{1}{4} + \dfrac{3}{8} + \dfrac{5}{16} + \dfrac{7}{32} + \dots$

$S = 1 \times\dfrac{1}{4} + 3 \times\dfrac{1}{8} + 5 \times\dfrac{1}{16} + 7 \times\dfrac{1}{32} + \dots$

This is a kind of arithmetic-o-geometric progression , which is made up of

1) A.P - $1 , 3 , 5 , 7 , \dots$ with (first term) $a = 1$ , $d = 2$

2) G.P - $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dots$ with (first term) $b = \dfrac{1}{4}$ , $r = \dfrac{1}{2}$

So , the sum of terms of such progressions for infinite terms is given by formula -

$S = \dfrac{ab}{1-r} + \dfrac{dbr}{(1-r)^2}$

Substituting the values of $a , b , d , r$ we get ,

$S = \dfrac{(1)\bigg(\dfrac{1}{4}\bigg)}{1 - \dfrac{1}{2}} + \dfrac{(2)\bigg(\dfrac{1}{4}\bigg)\bigg(\dfrac{1}{2}\bigg)}{\bigg(1 - \dfrac{1}{2}\bigg)^2}$

$S = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} + \dfrac{\dfrac{1}{4}}{\dfrac{1}{4}}$

$S = \dfrac{1}{2} + 1$

$S = \dfrac{3}{2}$

Hence , $a - b = 3 - 2 = \boxed{1}$

Note : This formula is applicable only if $| r | < 1$

First, note that for $n \ge 2$

$\frac{2(n-1)-1}{2^n} = \frac{2(n-1)+1}{2^{n-1}}-\frac{2n+1}{2^n}$

and each terms in the equations can be written as $\frac{2(n-1)-1}{2^n}$ for $n=2, 3, 4, 5, \dots$

So,

$\frac{1}{4}+\frac{3}{8}+\dots = \frac{3}{2}+(-\frac{5}{4}+\frac{5}{4}-\frac{7}{8}+\frac{7}{8}-\frac{9}{16}+\dots\infty)$

As the terms are summed to infinity, the sum in the bracket at $RHS$ will equal to $0$ . Therefore the entire sum is $\frac{3}{2}$ and the answer is $3-2=\boxed{1}$ .