A calculus problem by Rohith M.Athreya

Let $S_1$ be the required sum and $S_2 = \sum_{1 \leq i = j \leq n} \frac{1}{2^{i+j}}$

Now , $\sum_{1 \leq i \leq n} \sum_{1 \leq j \leq n} \frac{1}{2^{i+j}}$ . There will exist 3 cases, 2 of which correspond to $S_1$ while 1 will be $S_2$ . (Think of it as a matrix with $S_1$ being the sum of upper triangle or lower triangle and $S_2$ being the diagonal. Now the total sum itself can easily be calculated and $S_2$ also can easily be calculated to get the answer.)

$2S_1+S_2=\sum_{1 \leq i \leq n} \sum_{1 \leq j \leq n} \frac{1}{2^{i+j}}$ $=\sum_{1 \leq i \leq n} \frac{1}{2^i} \sum_{1 \leq j \leq n} \frac{1}{2^{j}}$ $=\sum_{1 \leq i \leq n} \frac{1}{2^i} (1)$ $=(1) (1)$ $=1$

$S_2=\sum_{1 \leq i = j \leq n} \frac{1}{2^{i+j}}$ $=\sum_{1 \leq i = \leq n} \frac{1}{2^{2i}}$ $=\frac{1/4}{1-\frac{1}{4}}$ $=\frac{1}{3}$

Thus the required answer is $2S_1=1-1/3$ or $S_1=1/3=0.333$

$\begin{aligned} S & = \lim_{n \to \infty} \sum_{1\le j < i \le n} \frac 1{2^{i+j}} \\ & = \underbrace{\sum_{\color{#3D99F6}i=2}^\infty \frac 1{2^{i+\color{#D61F06}1}}}_{\color{#D61F06}j=1} + \underbrace{\sum_{\color{#3D99F6}i=3}^\infty \frac 1{2^{i+\color{#D61F06}2}}}_{\color{#D61F06}j=2} + \underbrace{\sum_{\color{#3D99F6}i=4}^\infty \frac 1{2^{i+\color{#D61F06}3}}}_{\color{#D61F06}j=3} + \cdots & \small \color{#3D99F6} \text{Note that }i>j \\ & = \sum_{k=3}^\infty \frac 1{2^k} + \sum_{k=5}^\infty \frac 1{2^k} + \sum_{k=7}^\infty \frac 1{2^k} + \cdots \\ & = \frac 1{2^3} \sum_{k=0}^\infty \frac 1{2^k} + \frac 1{2^5} \sum_{k=0}^\infty \frac 1{2^k} + \frac 1{2^7} \sum_{k=0}^\infty \frac 1{2^k} + \cdots \\ & = \left( \frac 1{2^3} + \frac 1{2^5} + \frac 1{2^7} + \cdots \right)\sum_{k=0}^\infty \frac 1{2^k} \\ & = \frac 18 \left(1 + \frac 14 + \frac 1{4^2} + \cdots \right)\left(\frac 1{1-\frac 12}\right) \\ & = \frac 18 \left(\sum_{k=0}^\infty \frac 1{4^k} \right)\left(2\right) \\ & = \frac 14 \left(\frac 1{1-\frac 14} \right) \\ & = \frac 14 \left(\frac 43\right) = \frac 13 \approx \boxed{0.333} \end{aligned}$

$\displaystyle (\frac{1}{m}+\frac{1}{m^{2}}+.... infinite terms)^{2} = \frac{1}{m^{2}}+\frac{1}{m^{4}}+\frac{1}{m^{6}}+...infinite terms + 2\alpha$ where $\alpha$ is the sum asked for.

put m=2

what is done here is simply an algebraic multiplication extended to infinite terms