<s>[INST] Calculate the closed form of the following series $$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
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.
 [/INST] 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 $\displaystyle S = \frac{1}{3^r}$.
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