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I am trying to figure out what the following sum converges to:

$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$

An answer would be great, but if you have an explanation, that'd be better!

Archer
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ndm
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  • Have you meant $$\sum_{n=0}^\infty\binom{6+n}n x^n(6+n)$$ or $$\sum_{n=0}^\infty\binom{6+n}n x^{n(6+n)}$$ ? – lab bhattacharjee Feb 22 '14 at 04:23
  • $$\sum_{n=0}^\infty\binom{6+n}n x^n(6+n)$$ Thanks. – ndm Feb 22 '14 at 04:40
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    First of all,

    $$ {6+n\choose n}x^n(6+n)= {6+n\choose n}x^n(7+n-1)=(7+n)\frac{(6+n)!}{n! 6!}x^n-{6+n\choose n}x^n$$

    $$=\frac17{7+n\choose n}x^n-{6+n\choose n}x^n$$

    Now I was trying to find the relation between $$\sum_{n=1}^\infty{6+n\choose n}x^n$$

    and $$(1+x)^6+\frac{(1+x)^7}x+\frac{(1+x)^8}{x^2}+\cdots$$ which is an infinite Geometric Series

     – lab bhattacharjee Feb 22 '14 at 04:47
  • A not furthest relative : http://www.proofwiki.org/wiki/Rising_Sum_of_Binomial_Coefficients – lab bhattacharjee Feb 22 '14 at 05:07

4 Answers4

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Since $$ 7\binom{7+n}{n}=(7+n)\binom{6+n}{n} $$ and using negative binomial coefficients, $$ \binom{k+n}{n}=(-1)^n\binom{-k-1}{n} $$ we get $$ \begin{align} \sum_{n=0}^\infty\binom{6+n}{n}x^n(6+n) &=\sum_{n=0}^\infty\binom{6+n}{n}x^n(7+n)-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\ &=7\sum_{n=0}^\infty\binom{7+n}{n}x^n-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\ &=7\sum_{n=0}^\infty\binom{-8}{n}(-x)^n-\sum_{n=0}^\infty\binom{-7}{n}(-x)^n\\[3pt] &=\frac7{(1-x)^8}-\frac1{(1-x)^7}\\[6pt] &=\frac{6+x}{(1-x)^8} \end{align} $$

robjohn
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Related techniques: (I), (II). Follow the steps:

1) simplify $(n+6){ n+6\choose n} $ as

$$ (n+6){ n+6\choose n} = \frac{1}{6!}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2. $$

2) use the series identity

$$ \sum_{n=0}^{\infty} x^{n+1}=\frac{x}{1-x} \longrightarrow (*) $$

3) Applying the operators $D(xD)(x^2D)^5 $ to both sides of $(*)$ , where $D=\frac{d}{dx}$, gives

$$ \sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n+5}=D(xD)(x^2D)^5 \frac{x}{1-x} $$

$$ \implies \frac{1}{6!}\sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n}\\ =\frac{1}{6!\,x^5}D(xD)(x^2D)^5 \frac{x}{1-x} . $$

Note: The operator $(x^2 D)^5$ means

$$ (x^2 D)^5 = (x^2D)(x^2D)(x^2D)(x^2D)(x^2D). $$

Mhenni Benghorbal
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\left.\sum_{n = 0}^{\infty}{6 + n \choose n}x^{n}\pars{6 + n} \,\right\vert_{\ 0\ <\ x\ <\ 1}}} = {1 \over x^{5}}\sum_{n = 0}^{\infty}\bracks{{-7 \choose n}\pars{-1}^{n}} x^{n + 5}\pars{n + 6} \\[5mm] = &\ {1 \over x^{5}}\,\partiald{}{x} \sum_{n = 0}^{\infty}{-7 \choose n}\pars{-x}^{n + 6} = {1 \over x^{5}}\,\partiald{}{x}\bracks{\pars{-x}^{6} \sum_{n = 0}^{\infty}{-7 \choose n}\pars{-x}^{n}} = {1 \over x^{5}}\,\partiald{}{x}\bracks{x^{6}\pars{1 - x}^{\,-7}} \\[5mm] = &\ {1 \over x^{5}}\bracks{6x^{5}\pars{1 - x}^{-7} + 7x^{6}\pars{1 - x}^{-8}} = {6\pars{1 - x} + 7x \over \pars{1 - x}^{8}} = \bbx{6 + x \over \pars{1 - x}^{8}} \end{align}

Felix Marin
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Starting fom the previous answers to this post, it seems to me (assuming no mistakes o my side) that $$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n)=\frac{x+6}{(1-x)^8}$$

Claude Leibovici
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