Getting from $\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$ to $\frac{2}{3}\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{1}{3^m}$

Please see the following summation equality:

$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}=\frac{2}{3}\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{1}{3^m}$$

Can someone assist in understanding how to get from the LHS to the RHS? I understand the idea of factoring out $\frac{2}{3}$ but what happens to the $n$ in the numerator?

• In the sums do you mean $n=1$ instead of $n+1$? – Darth Geek Dec 31 '16 at 20:19
• On the right, use the formula for a geometric series. – Alex Dec 31 '16 at 20:23
• @DarthGeek yes my mistake. It's been corrected – ClownInTheMoon Dec 31 '16 at 20:25
• @Alex yes I see that this is an infinite geometric sequence but I'm more concerned with how the double summation is derived. – ClownInTheMoon Dec 31 '16 at 20:26
• @ClownInTheMoon I suggest that to show some effort, type up the first few steps (e.g., cancelling out the 2/3, applying the geometric series formula). The answer should be easy to deduce after that. – Alex Dec 31 '16 at 20:28

$$\sum_{n = 1}^\infty \frac{2n}{3^{n+1}} = \frac{2}{3}\sum_{n = 1}^\infty \frac{n}{3^n} \overset{(1)}{=} \frac{2}{3}\sum_{n = 1}^\infty \sum_{m = 1}^n \frac{1}{3^n} \overset{(2)}{=} \frac{2}{3}\sum_{m = 1}^\infty \sum_{n = m}^\infty \frac{1}{3^n} = \frac{2}{3}\sum_{n = 1}^\infty \sum_{m = n}^\infty \frac{1}{3^m}$$

In $(1)$, we use $n = \sum_{m = 1}^n 1$, and in $(2)$, we change the order of summation.

• (+1) Great to see you've returned. And Happy Holidays! -Mark – Mark Viola Dec 31 '16 at 20:42
• @Dr.MV thank you, and have a happy new year! – kobe Dec 31 '16 at 20:46

This is not a very rigorous explanation but hopefuly it will give you the idea behind this.

$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}= \frac{2}{3}\sum_{n=1}^\infty\frac{n}{3^n} = \frac{2}{3}\sum_{n=1}^\infty\sum_{k=1}^n\frac{1}{3^n} = \frac{2}{3}\sum_{n=1}^\infty\underbrace{\frac{1}{3^n} + \ldots + \dfrac{1}{3^n}}_{n} =$$

$$\dfrac{2}{3}\left[\left(\color{red}{\dfrac{1}{3^1}}\right) + \left(\color{blue}{\dfrac{1}{3^2}}+\color{blue}{\dfrac{1}{3^2}}\right)+\left(\color{green}{\dfrac{1}{3^3}}+\color{green}{\dfrac{1}{3^3}}+\color{green}{\dfrac{1}{3^3}}\right)+\left(\color{purple}{\dfrac{1}{3^4}} +\color{purple}{\dfrac{1}{3^4}} +\color{purple}{\dfrac{1}{3^4}} +\color{purple}{\dfrac{1}{3^4}} \right)+\ldots\right]$$

Reordering the terms, we have:

$$\dfrac{2}{3}\left[\left( \color{red}{\dfrac{1}{3^1}} + \color{blue}{\dfrac{1}{3^2}} + \color{green}{\dfrac{1}{3^3}} + \ldots \right) + \left( \color{blue}{\dfrac{1}{3^2}} + \color{green}{\dfrac{1}{3^3}} + \color{purple}{\dfrac{1}{3^4}} + \ldots \right) + \left( \color{green}{\dfrac{1}{3^3}} + \color{purple}{\dfrac{1}{3^4}} + \color{orange}{\dfrac{1}{3^5}} + \ldots \right) + \ldots\right]$$ which is precisely $\dfrac{2}{3}\displaystyle\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{1}{3^m}$.

(This can be done because the series is absolutely convergent).