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How do I solve sum of binomial series which is as follows: $$\frac{1}{3}\sum^{\infty}_{x=0}\begin{pmatrix}x\\y\end{pmatrix}\frac{1}{3}^x$$

I think it would be pretty easy to sum from $y=0$ to $x$, but I have no idea how to sum from $x=0$ to $x=\infty$.

Thanks,

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    $\begingroup$ Is it $\frac {1^x}3$ or $\big(\frac 13\big)^x$ $\endgroup$ – Claude Leibovici May 3 '16 at 3:48
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Remember binomial theorem: $$ (p+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y p^{x-y} $$ With $ p = 1/3 $: $$ (1/3+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y \left( \frac{1}{3} \right)^{x-y} = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} 3^y q^y \left( \frac{1}{3} \right)^{x} $$ Now summing over $ x $ on both sides and rearranging some things in the sum gives: $$ \sum_{x = 0}^\infty (1/3 + q)^x = \sum_{x} \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} 3^y q^y \left( \frac{1}{3} \right)^{x} \\ =\sum_{y} q^y 3^y \; \sum_{x} \begin{pmatrix}x\\y\end{pmatrix} \left( \frac{1}{3} \right)^{x} $$ So all you have to do to find your sum is to find the coefficient of $ q^y $ term in $ \sum_{x = 0}^\infty (1/3 + q)^x $, and divide by $ 3^y $. The sum is easily computed by geometric series: $$ \sum_{x = 0}^\infty (1/3 + q)^x = \frac{1}{2/3 - q} \\ = \frac{3}{2} \sum_y (3 q /2)^y $$ Where in the last line I used the geometric series expansion of $ 1/(1 - x) $ The coefficient of $ q^y $ is $(3/2)^{y+1} $, dividing by $ 3^y $ gives $ 3/ 2^{y+1} $.

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