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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...

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

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  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Mar 24 '16 at 4:56
  • $\begingroup$ 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" $\endgroup$ – user247327 Mar 24 '16 at 14:45
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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 $$\sum^\infty_{n=0} 2x^n +x^{2n} = \frac{2}{1-x} + \frac{1}{1-x^2}.$$

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