$$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
The answer given is $$\frac{1}{3^r}$$ I tried expanding the expression so it becomes $$\sum_{m=r}^{\infty}\frac{(m-1)!}{(r-1)!(m-r)!}\frac{1}{4^m}$$but I do not know how to follow.
Any help will be appreciated, thanks.
You can use recurrence relation, thanks to the Pascal's identity: $${m - 1 \choose k - 1} = {m \choose k} - {m - 1 \choose k}.$$
As suggested by Claude Leibovici in the comment we can have a more general result.
Let $$S(k) = \sum_{m = k}^\infty {m - 1 \choose k - 1} \frac{1}{x^m}.$$ It seems like $S(k)$ converges as long as $|x| > 1$.
Using Pascal's identity we have $$\sum_{m = k}^\infty {m - 1 \choose k - 1} \frac{1}{x^m} =x\sum_{m = k}^\infty {m \choose k} \frac{1}{x^{m + 1}} -\sum_{m = k}^\infty {m-1 \choose k} \frac{1}{x^{m}} $$ Which give us $S(k) = xS(k) - S(k - 1)$ or \begin{equation} S(k - 1) = (x - 1)S(k) \tag{1} \end{equation} Now $S(1)$ is just the geometric series $$\sum_{m = 1}^\infty \frac{1}{x^m} = \frac{1}{x - 1}$$ Which gives us the solution for $(1)$ is \begin{equation} S(k) = \frac{1}{(x - 1)^k} = \sum_{m = k}^\infty {m - 1 \choose k - 1} \frac{1}{x^m}\tag{2} \end{equation} Setting $x = 4$ gives us the desired result.
Note the similarity with negative binomial theorem: $$\frac{1}{(x + a)^k} = \sum_{j = 0}^\infty (-1)^j {k + j - 1 \choose j} x^j a^{k - j}$$ when $a = -1$ which converges for $|x| < 1$.
We can then combine $S(k)$ with this to get a (Laurent) series expansion of $\frac{1}{(x - 1)^k}$ as a corollary.
Say we have an unfair coin that yields head with probability $\frac{3}{4}$ and we toss it repeatedly until we get $r$ heads. The probability that the $r$-th head is from the $m$-th toss is given by the following expression:
$$ \binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m-r}\left(\frac{3}{4}\right)^{r} $$
If we sum the probability for all possible $m$ then we’ll get one
$$ \begin{align} 1&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m-r}\left(\frac{3}{4}\right)^{r}}\\ \\ &=3^{r}\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m}}\\ \\ \\ \text{or }&\text{equivalently}\\ \\ \\ \left(\frac{1}{3}\right)^{r}&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m}}\\ \end{align} $$
For a general case in which probability of getting head is $1-x$ where $0\leq x< 1$:
$$ \begin{align} 1&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}x^{m-r}\left(1-x\right)^{r}}\\ \\ &=\left(\frac{1-x}{x}\right)^{r}\sum_{m=r}^{\infty}{\binom{m-1}{r-1}x^{m}} \end{align} $$
We obtain \begin{align*} \color{blue}{\sum_{m=r}^\infty\binom{m-1}{r-1}\frac{1}{4^m}} &=\sum_{m=0}^\infty\binom{m+r-1}{m}\frac{1}{4^{m+r}}\tag{1}\\ &=\frac{1}{4^r}\sum_{m=0}^\infty\binom{-r}{m}\left(-\frac{1}{4}\right)^m\tag{2}\\ &=\frac{1}{4^r}\,\frac{1}{\left(1-\frac{1}{4}\right)^r}\tag{3}\\ &\,\,\color{blue}{=\frac{1}{3^r}} \end{align*} and the claim follows.
Comment:
In (1) we shift the index to start with $m=0$ and we also use the identity $\binom{p}{q}=\binom{p}{p-q}$.
In (2) we factor out $\frac{1}{4^r}$ and apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (3) we use the binomial series expansion.
Consider the series
$$S := \displaystyle \sum_{m=r}^{\infty}\binom{m-1}{r-1}x^m $$
We have the following factorial relation: $$\displaystyle \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$$
So that $\displaystyle \binom{m-1}{r-1} = \frac{r}{m}\binom{m}{r}. $ Hence we have
$$S := \sum_{m \ge r}\binom{m-1}{r-1}x^m = \sum_{m \ge r}\frac{r}{m}\binom{m}{r}x^m = \sum_{k \ge 0}\frac{r}{k+r}\binom{k+r}{r}x^{k+r} $$
For $\beta \in \mathbb C$ we have the binomial series $\displaystyle \frac{1}{(1-z)^{\beta+1}} = \sum_{k \ge 0} {k+\beta \choose k}z^k$.
Writing $\displaystyle \frac{1}{k+r} = \int_0^1 y^{k+r-1} \, \mathrm dy $ we have:
$$\begin{aligned} S & = rx^r\sum_{k \ge 0}\binom{k+r}{r}x^{k} \int_0^1 y^{r+k-1} \, \mathrm dy \\& = rx^r \int_0^1 \sum_{k \ge 0}\binom{k+r}{r}x^{k}y^{r+k-1} \, \mathrm dy \\& = rx^r \int_0^1 y^{r-1} \sum_{k \ge 0}\binom{k+r}{r}(xy)^k \, \mathrm dy \\& = rx^r \int_0^1 y^{r-1} \frac{1}{(1-xy)^{r+1}} \, \mathrm dy \\& = \frac{rx^r}{r(1-x)^r} \\& = \frac{x^r}{(1-x)^r}. \end{aligned} $$
The case where $\displaystyle x = \frac{1}{4}$ gives $\displaystyle S = \frac{1}{3^r}$.
Change the index of summation to $n:=m-1$. Your sum is then $$ \sum_{n=r-1}^\infty{n\choose r-1}\frac1{4^{n+1}}=\frac14\sum_{n=r-1}^\infty{n\choose r-1}\frac1{4^{n}}. $$ Now apply the identity $$\sum_{n=k}^\infty{n\choose k}x^n=\frac{x^k}{(1-x)^{k+1}} $$ with $k:=r-1$ and $x:=\frac14$ to obtain the result.
We can use the following identity (used to compute the power of an infinite Taylor series) directly to compute the sum:
$\left(\sum\limits_{k=1}^{\infty}x^k\right)^r=\sum\limits_{m=r}^{\infty}\binom{m-1}{r-1}{x^m}$
(refer to https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) (example 4)
We can understand the above equality combinatorially, by using stars and bars method.
Let's consider the number of ways to form $m^{th}$ power of $x$ on the RHS.
The above is equivalent to placing $m$ identical balls (RHS) into $r$ boxes (LHS), so that each box contains at least one ball
It can be done in $m-r+r-1 \choose r-1$= $ m-1 \choose r-1$ ways.
Now, choosing $x=\frac{1}{4}$, from the RHS of the above identity, we have,
$\sum\limits_{m=r}^{\infty}\binom{m-1}{r-1}{\left(\frac{1}{4}\right)^m}=\left(\sum\limits_{k=1}^{\infty}\left(\frac{1}{4}\right)^k\right)^r =\left(\frac{\frac{1}{4}}{1-\frac{1}{4}}\right)^r=\frac{1}{3^r}$
by using the formula for a sum of an infinite GP series.
