0
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ In the sums do you mean $n=1$ instead of $n+1$? $\endgroup$ – Darth Geek Dec 31 '16 at 20:19
  • $\begingroup$ On the right, use the formula for a geometric series. $\endgroup$ – Alex Dec 31 '16 at 20:23
  • $\begingroup$ @DarthGeek yes my mistake. It's been corrected $\endgroup$ – ClownInTheMoon Dec 31 '16 at 20:25
  • $\begingroup$ @Alex yes I see that this is an infinite geometric sequence but I'm more concerned with how the double summation is derived. $\endgroup$ – ClownInTheMoon Dec 31 '16 at 20:26
  • $\begingroup$ @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. $\endgroup$ – Alex Dec 31 '16 at 20:28
3
$\begingroup$

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

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

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

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).

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.