# Closed form for infinite sum $\;\displaystyle\sum \limits_{n=0}^{\infty} 2x^{n} + x^{2n}$ [closed]

Could someone help walk me through how to find a closed form for the sum $\;\,\displaystyle\sum \limits_{n=0}^{\infty} 2x^{n} + x^{2n}\;$? I ran into a sum of this form when trying to find the expected value for the maximum of two i.i.d. geometric random variables.

EDIT: Is it always the case that $\sum \limits_{x} f(x) + g(x) = \sum \limits_{x} f(x) + \sum \limits_{x} g(x)$? I feel like I knew this before but I forgot...

## closed as off-topic by user296602, Shahab, Stefan Mesken, Claude Leibovici, JonMark PerryMar 24 '16 at 8:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Shahab, Stefan Mesken, Claude Leibovici, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

• – lab bhattacharjee Mar 24 '16 at 4:56
• That is the same as $2\sum_{n=0}^\infty x^n+ \sum_{n=0}^\infty (x^2)^n$ both of which are "geometric series" – user247327 Mar 24 '16 at 14:45

## 1 Answer

We see $$\sum^\infty_{n=0} 2x^n +x^{2n} = \lim_{N\to \infty }\sum^N_{n=0} 2x^n +x^{2n} = 2\lim_{N\to \infty} \sum^N_{n=0}x^n ++ \lim_{N\to \infty} \sum^N_{n=0}x^{2n}$$ when the limits exist. So the first question is to decide when the limits exist. It turns out that both limit exist if and only if $\lvert x \rvert < 1$. In each case, we see $$\sum^N_{n=0} x^n = \frac{1-x^{N+1}}{1-x} \,\,\,\,\,\,\,\,\,\, \text{ and } \,\,\,\,\,\,\,\,\,\,\, \sum^N_{n=0} x^{2n} = \sum^N_{n=0} (x^{2})^n = \frac{1-(x^2)^{N+1}}{1-(x^2)};$$ indeed, the first can be proven easily by induction and the second follows directly from the first. Then since $\lvert x \rvert < 1$, we can take the limits to arrive at $$\sum^\infty_{n=0} 2x^n +x^{2n} = \frac{2}{1-x} + \frac{1}{1-x^2}.$$