# How is $\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}x^{i+j} = \sum_{n=0}^{\infty}\sum_{i=0}^{n}x^n$

My professor used this in class for a proof and I'm having trouble understanding it.

$$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}x^{i+j} = \sum_{n=0}^{\infty}\sum_{i=0}^{n}x^n$$

The way he explained it, you are reindexing the first summation using $$n = i+j$$, so $$0 \leq i \leq n$$ and then $$j = n-i$$, but this makes even less sense to me.

So my question is, what is going on here? How is this true? I'm totally lost.

(I have confirmed that this is true though https://www.wolframalpha.com/input/?i=sum+(+sum+(x%5En),+i+%3D+0+to+n),+n+%3D+0+to+infinity+%3D%3D%3D+sum+(+sum+(x%5E(i%2Bj)),+j+%3D+0+to+infinity),+i+%3D+0+to+infinity

• I don't think that is true.... – Lord Shark the Unknown May 15 at 18:50
• Note that this is only true if both double sums converge. It is possible for both to diverge, or one to diverge. – Don Thousand May 15 at 18:59

In the double sum $$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}x^{i+j}$$ $$x^n$$ appears when $$(i,j) \in \{ (0,n), (1,n-1),..., (n,0)\}$$, meaning $$n+1$$ times. Therefore $$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}x^{i+j} = \sum_{n=0}^{\infty}(n+1)x^n$$
Now, $$(n+1)x^n = \sum_{i=0}^{n}x^n$$
• Minor typo, $(n,0)$ instead of $(n\color{red}{-}0)$ – JMoravitz May 15 at 18:51
$$\begin{bmatrix} x^0 & x^1 & x^2 & x^3 \dots \\ 0 & x^1 & x^2 & x^3 \dots \\ 0 & 0 & x^2 & x^3 \dots \\ 0 & 0 & 0 & x^3 \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \\ \end{bmatrix}$$