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The question is to find out the value of $$\sum_{n=k}^{\infty} P^n \binom nk (1/2)^n$$

I tried to break down the binomial coefficient and bring it in form of some known sequence but could not get anything out of it.Please help me in this regard.Thanks.

Navin
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3 Answers3

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For $k \ge 1$, $-1 < x < 1$, \begin{equation*} (1-x)^{-k} = \sum_{n=0}^\infty \binom{n+k-1}{k-1} x^k. \end{equation*} You should be able to manipulate your sum into looking something like this, to get \begin{equation*} \sum_{n=k}^\infty P^n \binom{n}{k} (1/2)^n = \frac{P^k}{2^k (1 - P/2)^{k+1}}. \end{equation*} for $-2 < P < 2$.

Daniel Schepler
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = k}^{\infty}P^{n}{n \choose k}\pars{1 \over 2}^{n} & = \sum_{n = k}^{\infty}{n \choose n - k}\pars{P \over 2}^{n} = \sum_{n = k}^{\infty}\bracks{{-k - 1 \choose n - k}\pars{-1}^{n - k}} \pars{P \over 2}^{n} \\[5mm] & = \pars{P \over 2}^{k}\sum_{n = 0}^{\infty}{-k - 1 \choose n} \pars{-\,{P \over 2}}^{n} = \pars{P \over 2}^{k}\pars{1 - {P \over 2}}^{-k - 1} \\[5mm] & = \bbx{\ds{2\,P^{k}\,\pars{2 - P}^{-k - 1}}} \end{align}

Felix Marin
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$$\begin{align} \sum_{n=k}^{\infty}P^n{n\choose k}\left({1\over 2}\right)^n &= {P^k\over 2^kk!}\sum_{n=k}^\infty n(n-1)\dots(n-k+1)\left({P\over 2}\right)^{n-k} \\&= {P^k\over 2^kk!}f^{(k)}({P\over 2}) \end{align}$$ Where $f : x\mapsto \sum_{n=0}^\infty x^n = {1\over 1-x}$

Thus $$\sum_{n=k}^{\infty}P^n{n\choose k}\left({1\over 2}\right)^n = {P^k\over 2^kk!}{k!\over (1-P/2)^{k+1}} = {P^k\over 2^k(1-P/2)^{k+1}}$$

Astyx
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