Series of binomial coefficients: $\sum\limits_{n=k}^{\infty}{\binom nk}x^n=\frac{x^k}{(1-x)^{k+1}}$ Any hint to prove this? I tried with properties of binomial coefficient but I can't get anything.
$$\sum_{n=k}^{\infty}{{n}\choose{k}}x^n=\dfrac{x^k}{(1-x)^{k+1}}$$
 A: $\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}
$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$$
\begin{array}{rclr}
\ds{\color{#f00}{\sum_{n = k}^{\infty}{n \choose k}x^{n}}} & 
\ds{\stackrel{n\ \mapsto\ n + k}{=}} &\!\!\!\!\!
\ds{\sum_{n = 0}^{\infty}{n + k \choose k}x^{n + k}}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum_{n = 0}^{\infty}{n + k \choose n}x^{n}} &\pars{~Binomial\ Symmetry~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum_{n = 0}^{\infty}{\bracks{-n - k} + n - 1 \choose n}\pars{-1}^{n}x^{n}} &
\qquad\pars{~Negating\ Property~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum_{n = 0}^{\infty}{- k - 1 \choose n}\pars{-x}^{n} =
x^{k}\bracks{1 + \pars{-x}}^{-k - 1}} & \pars{~Binomial\ Theorem~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{\color{#f00}{x^{k} \over \pars{1 - x}^{k + 1}}} &
\end{array}
$$
A: You can prove it by induction on the formal power series too. The result is clear for $k=0$, in which case you have the geometric series $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$. 
For simplicity we will write the sum over the whole integer $n\in \mathbb{Z}$, with the classical convention (see generatingfunctionology
 by H. Wilf) that $\binom{n}{j}$ is $0$ if $j>n$ or $j < 0$.
If we take the result to be true for $k$, for $k+1$ we have
$$
(1-x)\sum_n \binom{n}{k+1} x^n = \sum_n \binom{n}{k+1}x^n - \sum_n \binom{n}{k+1}x^{n+1}
$$
and the second sum can be rewritten in terms of $n\mapsto n+1$ to get
$$
(1-x)\sum_n \binom{n}{k+1} x^n = \sum_n \binom{n}{k+1}x^n - \sum_n \binom{n-1}{k+1}x^n
$$
which means that the coefficients of $(1-x)\sum_n \binom{n}{k+1} x^n$ are
$$
[x^j] \left\{(1-x)\sum_n \binom{n}{k+1} x^n\right\} = \binom{j}{k+1}-\binom{j-1}{k+1}
$$
which equals $\binom{j-1}{k}$ by the classical Pascal's rule $\binom{j}{k+1}=\binom{j-1}{k+1}+\binom{j-1}{k}$.
Thus
$$
(1-x)\sum_n \binom{n}{k+1} x^n = \sum_n \binom{n-1}{k}x^n = x\sum_n \binom{n}{k}x^n = x\frac{x^{k}}{(1-x)^{k+1}}\,,
$$
where we have set $n\mapsto n-1$ in the second equality, and used the inductive hypothesis in the third. Now the result follows upon dividing by $1-x$.
Moral of the story Intuitively, the identity $$\sum_n \binom{n}{k}x^n=\frac{x^{k}}{(1-x)^{k+1}}$$ is a substractive application of Pascal's rule, while you can try and see that $$\sum_k \binom{n}{k}x^k = (1+x)^n$$ would be the additive one.
A: 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 $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$
