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.

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.
