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

Question

Any hint to prove this? I tried with properties of binomial coefficient but I can't get anything.

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

Answer 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 $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$

Answer 2

$\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} $

Answer 3

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

Answer 4

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.

Answer 5

The probability that you obtain the $(k+1)$-th success after $n+1$ Bernoulli trials with failure probability $x$ is $\binom nk(1-x)^{k+1}x^{n-k}$, since the first $k$ successes can be any of the first $n$ trials. Thus

$$ \sum_{n=k}^\infty\binom nk(1-x)^{k+1}x^{n-k}=1\;, $$

or

$$ \sum_{n=k}^\infty\binom nkx^n=\frac{x^k}{(1-x)^{k+1}}\;. $$

Answer 6

Here is a self-contained proof based on the idea of combinatorial class.

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Note the combinatorial interpretation of the binomial coefficient:

\begin{align*} \binom{n}{k} &= [\text{# of length-$n$ strings in alphabet $\{\texttt{a}, \texttt{b}\}$ containing exactly $k$ $\texttt{b}$'s}] \end{align*}

So, if $w$ stands for any string in alphabet $\{\texttt{a}, \texttt{b}\}$ and $|w|$ denotes the length of $w$, then

$$ \binom{n}{k} = \sum_{\substack{|w| = n \\ \text{$w$ has exactly $k$ $\texttt{b}$'s}}} 1 $$

and hence

\begin{align*} \sum_{k=n}^{\infty} \binom{n}{k} x^n &= \sum_{k=n}^{\infty} \sum_{\substack{|w| = n \\ \text{$w$ has exactly $k$ $\texttt{b}$'s}}} x^n \\ &= \sum_{k=n}^{\infty} \sum_{\substack{|w| = n \\ \text{$w$ has exactly $k$ $\texttt{b}$'s}}} x^{|w|} \\ &= \sum_{\text{$w$ has exactly $k$ $\texttt{b}$'s}} x^{|w|}. \end{align*}

On the other hand, any word $w$ in alphabet $\{\texttt{a}, \texttt{b}\}$ having exactly $k$ $\texttt{b}$'s can be parametrized in the form

$$ \texttt{a}^{i_1}\texttt{b}\texttt{a}^{i_2}\texttt{b}\cdots \texttt{b}\texttt{a}^{i_{k+1}}, $$

where $\texttt{a}^0 = \text{(empty string)}$, $\texttt{a}^1= \texttt{a}$, $\texttt{a}^2 = \texttt{aa}$, $\texttt{a}^3 = \texttt{aaa}$, etc. Consequently, the last sum can be expanded as

\begin{align*} &= \sum_{i_1 = 0}^{\infty} \cdots \sum_{i_{k+1} = 0}^{\infty} x^{i_1+i_2+\cdots+i_{k+1} + k} \\ &= x^k \biggl( \sum_{i_1 = 0}^{\infty} x^{i_1} \biggr) \cdots \biggl( \sum_{i_{k+1} = 0}^{\infty} x^{i_{k+1}} \biggr) \\ &= \frac{x^k}{(1-x)^{k+1}}. \end{align*}