How can I find $\sum\limits_{n=i+1}^\infty \binom{n-1}{i}\left (\frac{1}{3}\right)^{n}$? In a probability proof I've arrived at the sum $\sum\limits_{n=i+1}^\infty \binom{n-1}{i} \left(\frac{1}{3}\right)^{n}$ where $i$ is constant. WolframAlpha gives a simple expression for this sum, but I've been unable to find it myself: I've tried rewriting it as $\frac{1}{i!}\sum\limits_{n=i+1}^\infty \frac{(n-1)!}{(n-1-i)!} \left(\frac{1}{3}\right)^{n}$ but this hasn't advanced me much.
How can I find this sum?

Using Grigory's Wiki link, we wish to adjust our sum to start at zero.
Find $m$ such that $n = i+1$ implies $m=0$: $m = n-(i+1) = n-i-1.$ Now adjust the term inside the binomial coefficient. Since $n = m+i+1$, then $n-1 = m+i$.
Therefore,
$$\sum_{n=i+1}^\infty \begin{pmatrix} n-1 \\ i\end{pmatrix} \left(\frac13\right)^n = \sum_{m=0}^\infty \begin{pmatrix} m+i \\ i \end{pmatrix}\left(\frac13\right)^{m+i+1}.$$
Now, $i$ is fixed inside the sum. So let's re-write:
$$\sum_{m=0}^\infty \begin{pmatrix} m+i \\ i \end{pmatrix}\left(\frac13\right)^{m+i+1} = \left(\frac13\right)^{i+1}\sum_{m=0}^\infty \begin{pmatrix} m+i \\ i \end{pmatrix}\left(\frac13\right)^m.$$
The binomial coefficient is symmetric, in other words
$$\begin{pmatrix} p \\ q \end{pmatrix} = \frac{p!}{q!(p-q)!} = \frac{p!}{(p-q)!(p-(p-q))!} = \begin{pmatrix} p \\ p-q\end{pmatrix}.$$
Therefore, $$\begin{pmatrix} m+i \\ i\end{pmatrix} = \begin{pmatrix} m+i \\ m+i-i \end{pmatrix} = \begin{pmatrix} m+i \\ m \end{pmatrix}.$$
Now our series becomes
$$\begin{align*}
\left(\frac13\right)^{i+1}\sum_{m=0}^\infty \begin{pmatrix} m+i \\ i \end{pmatrix}\left(\frac13\right)^m &= \left(\frac13\right)^{i+1}\sum_{m=0}^\infty \begin{pmatrix} m+i \\ m \end{pmatrix}\left(\frac13\right)^{m} \\
 &= \frac{\left(\frac13\right)^{i+1}}{\left(1-\frac13\right)^{i+1}}\\
 &= \frac{\left(\frac13\right)^{i+1}}{\left(\frac23\right)^{i+1}} \\
 &= 2^{-i-1}.
\end{align*}$$
Note that the second step uses the special case linked in the Wiki article.
