Determine infinite sum with shifting binomial coefficients

Question

I am looking for a way to find analytically the following sum

$\sum_{i=0}^\infty \frac{1}{6^i} \begin{pmatrix} i+k\\ i \end{pmatrix} = \Big(\frac65\Big)^{1+k}$,

for some integer $k>0$. The answer I displayed here was found by Wolfram Alpha. I looked at partial sums, but it involves hypergeometric functions. Is there any way around this? Many thanks in advance.

Answer 1

Note that $$F\left(k+1\right)=1+\sum_{i\geq1}\dbinom{i+k+1}{i}\frac{1}{6^{i}} $$ $$=1+\sum_{i\geq1}\left(\dbinom{i+k}{i}+\dbinom{i+k}{i-1}\right)\frac{1}{6^{i}} $$ $$=1+\sum_{i\geq1}\dbinom{i+k}{i}\frac{1}{6^{i}}+\sum_{i\geq1}\dbinom{i+k}{i-1}\frac{1}{6^{i}} $$ $$=\sum_{i\geq0}\dbinom{i+k}{i}\frac{1}{6^{i}}+\frac{1}{6}\sum_{i\geq0}\dbinom{i+k+1}{i}\frac{1}{6^{i}} $$ $$=F\left(k\right)+\frac{F\left(k+1\right)}{6} $$ hence $$F\left(k+1\right)=\frac{6}{5}F\left(k\right) $$ and since $$F\left(1\right)=\sum_{i\geq0}\dbinom{i+1}{i}\frac{1}{6^{i}}=\sum_{i\geq0}\frac{i+1}{6^{i}}=\left(\frac{6}{5}\right)^{2} $$ we have $$F\left(k+1\right)=\left(\frac{6}{5}\right)^{k+2}. $$

Answer 2

Consider the Taylor expansion at $x=0$ of $$\frac{1}{(1-x)^{1+k}}=\sum_{i=0}^{\infty}\binom{-(1+k)}{i}(-x)^i$$ which holds for $|x|<1$. Then note that $$\binom{-(1+k)}{i}=\frac{(-(1+k))\cdot (-(1+k)-1)\cdot (-(1+k)-i+1)}{i!}=(-1)^i\binom{i+k}{i}.$$ Hence $$\sum_{i=0}^\infty \frac{1}{6^i} \binom{i+k}{i}=\frac{1}{(1-(1/6))^{1+k}} = \Big(\frac65\Big)^{1+k}.$$

Answer 3

$\newcommand{\bbx}[1]{\bbox[8px,border:1px groove navy]{{#1}}\ } \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = 0}^{\infty}{1 \over 6^{i}}{i + k \choose i} & = \sum_{i = 0}^{\infty}{1 \over 6^{i}} {-\bracks{i + k} + i - 1 \choose i}\pars{-1}^{i}\qquad \pars{~Binomial\ Negating\ Property~} \\[5mm] & = \sum_{i = 0}^{\infty}{-k - 1 \choose i}\pars{-\,{1 \over 6}}^{i} = \bracks{1 + \pars{-\,{1 \over 6}}}^{-k - 1} =\ \bbx{\pars{6 \over 5}^{k + 1}} \end{align}