# Infinite sum involving binomial coefficients and exponentials

I found out this interesting sum while solving a series problem: $$\sum_{n=3}^{\infty}\binom n3\cdot5^{-n}$$ and by solving the series by another method (written below), I got the sum to be $$\dfrac{5}{256}$$. Nonetheless, if this problem was given in this form, I most likely wouldn't be able to convert it into a easily-dealt-with form. My question is, what are the other ways to solve this sum apart from converting into an AGP of a bunch of AGPs, which isn't quite easily motivated?

Background

I had this problem: $$\begin{equation*} \begin{split} S&=1\left(5^{-3}+2\cdot5^{-4}+3\cdot5^{-5}+\cdots\right)\\ &+2\left(5^{-4}+2\cdot5^{-5}+3\cdot5^{-6}+\cdots\right)\\ &+3\left(5^{-5}+2\cdot5^{-6}+3\cdot5^{-7}+\cdots\right)\\ &+\cdots \end{split} \end{equation*}$$ And I solved it by noticing that by suitable manipulations for each particular sum, we can reduce it to the form $$1+2x+3x^2+\cdots=\dfrac{1}{(1-x)^2}$$ which can be solved by standard method of solving AGPs, or by differentiating the geometric series.

Thus, the sum can be written as: $$S=5^{-3}\cdot\dfrac{1}{(1-5^{-1})^2}+2\cdot5^{-4}\cdot\dfrac{1}{(1-5^{-1})^2}+3\cdot5^{-5}\cdot\dfrac{1}{(1-5^{-1})^2}+\cdots$$ which in turn is again of the same form and thus is $$S=\dfrac{5^{-3}}{(1-5^{-1})^2}\cdot\dfrac{1}{(1-5^{-1})^2}=\dfrac{5}{256}$$

But if we rather group the terms of the form $$5^{\textrm{something}}$$ then we have $$S=(1\cdot 1)5^{-3}+(1\cdot2+2\cdot 1)5^{-4}+(1\cdot3+2\cdot2+3\cdot1)5^{-5}+\cdots$$ which by observing the pattern is nothing other than $$S=\sum_{n=3}^{\infty}\binom n3\cdot5^{-n},$$ the original sum.

By Newton's binomial theorem, $$\sum_{n=k}^\infty \binom{n}{k} z^n =z^k\sum_{n=0}^\infty \binom{n+k}{k} z^n =\frac{z^k}{(1-z)^{k+1}}$$ Now take $$k=3$$ and $$z=1/5$$ to obtain $$\sum_{n=3}^\infty \binom{n}{3} 5^{-n} =\frac{(1/5)^3}{(4/5)^4}=\frac{5}{256}$$