How to prove $\frac{1}{1-p}=\sum_{n=r}^\infty {{n \choose r}p^{n-r}(1-p)^r }$

Question

As we know $\frac{1}{(1-p)^{r+1}}=\sum_{k=0}^\infty{{k+r \choose k} p^k}$ and $\frac{1}{1-p}=\sum_{k=0}^\infty {p^k}$. But how to prove $$\frac{1}{1-p}=\sum_{n=r}^\infty {{n \choose r}p^{n-r}(1-p)^r }$$

Answer 1

This can be derived easily from the first equation you presented. Notice that

$$ \frac{1}{(1-p)^{r+1}} \;\; =\;\; \sum_{k=0}^\infty \binom{k+r}{k} p^k \;\;\;\; \Longrightarrow \;\;\;\; \frac{1}{1-p} \;\; =\;\; \sum_{k=0}^\infty\binom{k+r}{k}p^k (1-p)^r. $$

Notice now that we take the substitution $n = k+r$. This forces the sum to start when $n = r$, thus the sum becomes

$$ \frac{1}{1-p} \;\; =\;\; \sum_{n=r}^\infty \binom{n}{n-r} p^{n-r} (1-p)^r. $$

The last part follows from the fact that

$$ \binom{n}{n-r} \;\; =\;\; \frac{n!}{r!(n-r)!} \;\; =\;\; \binom{n}{r}. $$

Answer 2

Of course, we need $|p|<1$ (for uniform convergence, that allows us to differentiate term by term). Write $f(t)=1/(1-t)$. We have \begin{align} \sum_{n=r}^\infty {{n \choose r}p^{n-r}(1-p)^r } &=(1-p)^r \,\sum_{n=r}^\infty {{n \choose r}p^{n-r}} =\frac{(1-p)^r}{r!} \,\sum_{n=r}^\infty n(n-1)\cdots(n-r+1)\,p^{n-r}\\ \ \\ &=\frac{(1-p)^r}{r!} \,f^{(r)}(p) =\frac{(1-p)^r}{r!} \,\frac{r!}{(1-p)^{r+1}}=\frac1{1-p}.\\ \end{align}

Answer 3

This is a consequence of

$$ \sum_{n=k}^\infty {n\choose k} y^n = \frac{y^k}{(1-y)^{k+1}}\;, $$

the generating function of binomial coefficients with fixed lower index. This you can show by differentiating

$$ \sum_{n=0}^\infty y^{n}=(1-y)^{-1} $$

$k$ times with respect to $y$ and then multiplying through by $y^k$.

Answer 4

Here we present a "brute-force" way forward. First, we translate the summation index enforcing the substitution $n\to n+r$ on the series of interest $S_r$ to obtain

$$\begin{align} S_r&=\sum_{n=r}^\infty \binom{n}{r}p^{n-r}(1-p)^r\\\\ &=(1-p)^{r}\sum_{n=0}^\infty \binom{n+r}{r}p^n \tag 1\\\\ \end{align}$$

Now, bringing a factor of $1-p$ inside the series in $(1)$ reveals

$$\begin{align} S_r&=(1-p)^{r-1}\sum_{n=0}^\infty \binom{n+r}{r}\left(p^n-p^{n+1}\right)\\\\ &=(1-p)^{r-1}\left(1+\sum_{n=1}^\infty \left(\binom{n+r}{r}-\binom{n+r-1}{r}\right)p^n\right)\\\\ &=(1-p)^{r-1}\sum_{n=0}^\infty \binom{n+r-1}{r-1}p^n\\\\ &=S_{r-1} \end{align}$$

Since $S_r=S_{r-1}$, then by induction $S_r=S_0$ and we find that

$$\begin{align} S_0&=\sum_{n=0}^\infty p^n\\\\ &=\frac{1}{1-p} \end{align}$$

as was to be shown! And we are done.

Answer 5

Edit: (some changes due to a remark of @NF Dream) As you mention probabilities in your tags, there is a probabilistic proof using the fact that for $X$ a Random Variable having a negative binomial distribution with parameter $p$ and $r$ (see https://math.stackexchange.com/q/1071294), we have:

$$P(X=n)= {{n \choose n-r}p^{n-r}(1-p)^{r+1}} \ \ \ n=r,r+1,...$$

and then applying the fact that $\sum_{n=r}^{+\infty}P(X=n)=1$, and terminating by dividing both sides by $1-p$.

Answer 6

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{\quad 0\ \leq\ p\ <\ 1\quad}$ and $\ds{\quad -1\ <\ z\ <\ 1}$, we'll use a "Generating Function Method":

\begin{align} \sum_{r = 0}^{\infty}z^{r}\bracks{\sum_{n = r}^{\infty}{n \choose r} p^{n - r}\pars{1 - p}^{r}} & = \sum_{n = 0}^{\infty}p^{n}\bracks{\sum_{r = 0}^{n}{n \choose r} \pars{z\,{1 - p \over p}}^{r}} = \sum_{n = 0}^{\infty}p^{n}\pars{1 + z\,{1 - p \over p}}^{n} \\[3mm] & = \sum_{n = 0}^{\infty}\bracks{p + z\pars{1 - p}}^{\,n}= {1 \over 1 - \bracks{p + z\pars{1 - p}}} = {1 \over 1 - p}\,{1 \over 1 - z} \end{align}

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Then, $$ \sum_{r = 0}^{\infty}z^{r}\bracks{\sum_{n = r}^{\infty}{n \choose r} p^{n - r}\pars{1 - p}^{r}} = \sum_{r = 0}^{\infty}z^{r}\bracks{{1 \over 1 - p}} \ \imp\ \fbox{$\ds{\quad% \sum_{n = r}^{\infty}{n \choose r}p^{n - r}\pars{1 - p}^{r} = {1 \over 1 - p}\quad}$} $$