Series of binomial coefficients: $\sum\limits_{n=k}^{\infty}{\binom nk}x^n=\frac{x^k}{(1-x)^{k+1}}$

For $|x|<1,$ we have $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}.$$ Now take $k$ derivatives on both sides: $$\sum_{n=k}^\infty\frac{n!}{(n-k)!}x^{n-k}=\frac{k!}{(1-x)^{k+1}}.$$ Thus, $$\sum_{n=k}^\infty\frac{n!}{(n-k)!k!}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$ Since $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$

$\sum_{n=k}^{\infty}{{n}\choose{k}}x^n=\dfrac{x^k}{(1-x)^{k+1}} $

Start with $\sum_{n=k}^{\infty}{{n}\choose{k}}x^n =\sum_{n=0}^{\infty}{{n}\choose{n+k}}x^{n+k} =x^k\sum_{n=0}^{\infty}{{n+k}\choose{k}}x^{n} $

and note that $(1-x)^{-m} =\sum_{k=0}^{\infty} \binom{-m}{k} x^k $ and

$\begin{array}\\ \binom{-m}{k} &=\dfrac{\prod_{j=0}^{k-1}(-m-j)}{k!}\\ &=(-1)^k\dfrac{\prod_{j=0}^{k-1}(m+j)}{k!}\\ &=(-1)^k\dfrac{\prod_{j=0}^{k-1}(m+k-1-j)}{k!}\\ &=(-1)^k\dfrac{(m+k-1)!}{(m-1)!k!}\\ &=(-1)^k\binom{m+k-1}{k}\\ \end{array} $

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \begin{array}{rclr} \ds{\color{#f00}{\sum_{n = k}^{\infty}{n \choose k}x^{n}}} & \ds{\stackrel{n\ \mapsto\ n + k}{=}} &\!\!\!\!\! \ds{\sum_{n = 0}^{\infty}{n + k \choose k}x^{n + k}} \\[5mm] & \ds{=} &\!\!\!\!\! \ds{x^{k}\sum_{n = 0}^{\infty}{n + k \choose n}x^{n}} &\pars{~Binomial\ Symmetry~} \\[5mm] & \ds{=} &\!\!\!\!\! \ds{x^{k}\sum_{n = 0}^{\infty}{\bracks{-n - k} + n - 1 \choose n}\pars{-1}^{n}x^{n}} & \qquad\pars{~Negating\ Property~} \\[5mm] & \ds{=} &\!\!\!\!\! \ds{x^{k}\sum_{n = 0}^{\infty}{- k - 1 \choose n}\pars{-x}^{n} = x^{k}\bracks{1 + \pars{-x}}^{-k - 1}} & \pars{~Binomial\ Theorem~} \\[5mm] & \ds{=} &\!\!\!\!\! \ds{\color{#f00}{x^{k} \over \pars{1 - x}^{k + 1}}} & \end{array} $$