This requires intuitive approach!

Let us fix $i+j=k$ for a certain $k$ ranging between $0$ and $\infty$ . Now the question is: How many way we can get the sum $i+j$ to be $k$ ?

Note that for $i+j=k$ to be fulfilled, $i$ and $j$ should range between $0$ and $k$ : if $i=0 , j=k$ , if $i=1 , j=k-1$ ... And thus $k+1$ possibilities. Therefore:

$\displaystyle \frac{2^{-(i+j)}}{i+j+1} = (k+1) \frac{2^{-(k)}}{k+1} = 2^{-k}$

And since $(i,j )=[0,\infty]\times [0,\infty] => k \in [0,\infty]$ Therefore our sum is equivalent to :

$\displaystyle \sum_{k=0}^{\infty} 2^{-k} = \boxed{2}$

An interesting observation is that the term only depends on $i+j$ , which makes us wish to be able to rearrange the terms of the series. However, this is not always possible (for example, rearranging the terms of a conditionally convergent series might change the sum due to Riemann series theorem ). So we need to establish that the series we have is unconditionally convergent (rearranging the terms don't change the sum).

Since $\mathbb{R}$ (with the usual Euclidean metric) is complete, it is enough to show that the series is absolutely convergent. Since all terms of the series are non-negative, it is enough to show that the series converges.

We invoke several theorems:

• If $\{a_i\}, \{b_i\}$ are two sequences of real numbers where $a_i \le b_i$ for all $i$ , then $\sum a_i \le \sum b_i$ provided both series converge.
• If $\{a_i\}$ is an non-decreasing sequence of real numbers that is bounded, then $\{a_i\}$ converges.

As a corollary, if $\{a_i\}, \{b_i\}$ are two sequences of non-negative real numbers where $a_i \le b_i$ for all $i$ , and $\sum b_i$ converges, then $\sum a_i$ converges as well. (Observe the partial sums of $\{a_i\}$ ; this is bounded by $\sum b_i$ by the first theorem. The partial sums thus converge by the second theorem; this is $\sum a_i$ .)

In this case, we will use this corollary twice.

First, observe that $\frac{2^{-(i+j)}}{i+j+1} \le 2^{-(i+j)}$ for all $i,j \ge 0$ . Moreover, note that the series $\sum_{j=0}^\infty 2^{-(i+j)}$ converges for any $i$ (it's just a geometric series, with sum $2^{-i+1}$ ). Thus, by the corollary, the series $\sum_{j=0}^\infty \frac{2^{-(i+j)}}{i+j+1}$ also converges; let's call the value $b_i$ . We know it satisfies $0 \le b_i \le 2^{-i+1}$ (the left-hand inequality is because the terms are non-negative and so the sum is non-negative, the right-hand inequality is due to the first theorem).

Next, we observe the sequence $\{b_i\}_{i=0}^\infty$ , which is the thing we want to sum. We know that $b_i \le 2^{-i+1}$ for all $i$ . Additionally, the series $\sum_{i=0}^\infty 2^{-i+1}$ converges (to $2^2 = 4$ , geometric series again), so the series $\sum_{i=0}^\infty b_i$ also converges. This means the entire thing converges, and thus is absolutely convergent, and thus is unconditionally convergent. Hence we can rearrange the terms at will.

Now, observe that all terms only depend on $i+j$ , and not what $i,j$ themselves are. Thus we can collect terms having equal $i+j$ . For any non-negative integer $k$ , there are $k+1$ pairs of $(i,j)$ adding to $k$ (namely $(0,k), (1,k-1), (2,k-2), \ldots, (k,0)$ ), all appearing in the series. Hence, instead of working with $i,j$ , we can work on $k$ instead:

$\displaystyle\begin{aligned} \sum_{i=0}^\infty \sum_{j=0}^\infty \frac{2^{-(i+j)}}{i+j+1} &= \sum_{k=0}^\infty (k+1) \cdot \frac{2^{-k}}{k+1} \\ &= \sum_{k=0}^\infty 2^{-k} \\ &= 2^1 = \boxed{2} \end{aligned}$

$I=\sum _{ j=0 }^{ \infty }{ \sum _{ i=0 }^{ \infty }{ \frac { { 2 }^{ -(i+j) } }{ i+j+1 } } } =\quad \sum _{ j=0 }^{ \infty }{ \sum _{ i=0 }^{ \infty }{ { 2 }^{ -(i+j) } } } .\int _{ 0 }^{ 1 }{ { t }^{ i+j } } dt\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad I\quad =\int _{ 0 }^{ 1 }{ { \sum _{ j=0 }^{ \infty }{ \sum _{ i=0 }^{ \infty }{ { \left( \frac { t }{ 2 } \right) }^{ i+j } } } } } dt\\ Now\quad this\quad is\quad nothing\quad but\quad integral\quad of\quad infinite\quad summation\quad of\quad gp.\\ Therefore,\quad I\quad =\quad 2$

I constructed the folowing table, which made me see the answer right away:

$\begin{array}{c c c c c c} \textbf{i,j} & \textbf{0} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4}\ldots \\ \textbf{0}&1&\textcolor{#D61F06}{\dfrac{2^{-1}}{2}}&\dfrac{2^{-2}}{3}&\textcolor{#D61F06}{\dfrac{2^{-3}}{4}}&\dfrac{2^{-4}}{5} \ldots\\ \textbf{1}&\textcolor{#D61F06}{\dfrac{2^{-1}}{2}}&\dfrac{2^{-2}}{3}&\textcolor{#D61F06}{\dfrac{2^{-3}}{4}}&\dfrac{2^{-4}}{5}&\textcolor{#D61F06}{\dfrac{2^{-5}}{6}} \ldots\\ \textbf{2}&\dfrac{2^{-2}}{3}&\textcolor{#D61F06}{\dfrac{2^{-3}}{4}}&\dfrac{2^{-4}}{5}&\textcolor{#D61F06}{\dfrac{2^{-5}}{6}}&\dfrac{2^{-6}}{7} \ldots\\ \textbf{3}&\textcolor{#D61F06}{\dfrac{2^{-3}}{4}}&\dfrac{2^{-4}}{5}&\textcolor{#D61F06}{\dfrac{2^{-5}}{6}}&\dfrac{2^{-6}}{7}&\textcolor{#D61F06}{\dfrac{2^{-7}}{8}} \ldots\\ \textbf{4}&\dfrac{2^{-4}}{5}&\textcolor{#D61F06}{\dfrac{2^{-5}}{6}}&\dfrac{2^{-6}}{7}&\textcolor{#D61F06}{\dfrac{2^{-7}}{8}}&\dfrac{2^{-8}}{9} \ldots\\ \vdots &\vdots&\vdots&\vdots&\vdots&\vdots \quad \ddots \end{array}$

$\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\dfrac{2^{-i-j}}{i+j+1}=1+2\times\dfrac{2^{-1}}{2}+3\times\dfrac{2^{-2}}{3}+\ldots=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots=\dfrac{1}{1-\frac{1}{2}}=\boxed{2}$ .

so do the incredible substitution x = i + j. for each x, we end up having x + 1 pairs (i, j) such that i + j = x, so we can really just say this is \sum {k = 0}^{\infty} \frac{2^{-k}}{x + 1} * (x + 1) = \sum {k = 0}^{\infty} 2^{-k} = 2.