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?
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$\ds{\sum_{n\ =\ k + 1}^{\infty}{n - 1 \choose k}\pars{1 \over 3}^{n}:\
{\large ?}}$

$\Large\left.1\right)$
$$
\mbox{We'll use the identity}\quad
\bbox[10px,border:1px dotted black]{\ds{{s \choose \ell} =
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}}}
$$

\begin{align}
&\sum_{n\ =\ k + 1}^{\infty}{n - 1 \choose k}\pars{1 \over 3}^{n}
=\sum_{n\ =\ k + 1}^{\infty}\bracks{%
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}
\pars{1 \over 3}^{n}
\\[3mm]&={1 \over 3}\oint_{\verts{z}\ =\ 1}{1 \over z^{k + 1}}\bracks{%
\sum_{n\ =\ k + 1}^{\infty}{\pars{1 + z  \over 3}^{n - 1}}}\,{\dd z \over 2\pi\ic}
\\[3mm]&={1 \over 3}\oint_{\verts{z}\ =\ 1}{1 \over z^{k + 1}}\bracks{%
{\pars{1 + z}^{k}/3^{k} \over 1 - \pars{1 + z}/3}}\,{\dd z \over 2\pi\ic}
={1 \over 3^{k}}\,\half\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{k} \over z^{k + 1}}
{1 \over 1 - z/2}\,{\dd z \over 2\pi\ic}
\\[3mm]&={1 \over 3^{k}}\,\half\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{k} \over z^{k + 1}}\sum_{n = 0}^{\infty}\pars{z \over 2}^{n}
\,{\dd z \over 2\pi\ic}
={1 \over 3^{k}}\,\half\sum_{n = 0}^{\infty}{1 \over 2^{n}}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{k} \over z^{k - n+ 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&={1 \over 3^{k}}\,\half\sum_{n = 0}^{k}{1 \over 2^{n}}{k \choose k - n}
={1 \over 3^{k}}\,\half\sum_{n = 0}^{k}{k \choose n}\pars{1 \over 2}^{n}
={1 \over 3^{k}}\,\half\pars{1 + \half}^{k}
={1 \over 2^{k + 1}}
\end{align}
$$
\bbox[10px,border:1px dotted black]{\ds{%
\sum_{n\ =\ i + 1}^{\infty}{n - 1 \choose i}\pars{1 \over 3}^{n} = 
{1 \over 2^{i + 1}}}}
$$


$\Large\left.2\right)$
\begin{align}
\sum_{n\ =\ i + 1}^{\infty}{n - 1 \choose i}\pars{1 \over 3}^{n} & =
\sum_{n\ =\ 0}^{\infty}{n + i \choose i}\pars{1 \over 3}^{n + i + 1} =
\pars{1 \over 3}^{i + 1}
\sum_{n\ =\ 0}^{\infty}{n + i \choose n}\pars{1 \over 3}^{n}
\\[5mm] & =
\pars{1 \over 3}^{i + 1}
\sum_{n\ =\ 0}^{\infty}\bracks{{-i - 1 \choose n}\pars{-1}^{n}}
\pars{1 \over 3}^{n}
\\[5mm] & =
\pars{1 \over 3}^{i + 1}
\sum_{n\ =\ 0}^{\infty}{-i - 1 \choose n}\pars{-\,{1 \over 3}}^{n} =
\pars{1 \over 3}^{i + 1}\bracks{1 + \pars{-\,{1 \over 3}}}^{-i - 1}
\\[5mm] & =
\bbox[10px,border:1px dotted black]{\ds{1 \over 2^{i + 1}}}
\end{align}
A: 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.
A: This is a generating function approach.
Let $$S_i(z)=\sum_{n=i}^\infty \binom n i z^i$$
The number you are looking for is $\frac{1}{3}S_i\left(\frac 1 3\right)$.
Let $$\begin{align}
T(w,z) &= \sum_{i=0}^\infty S_i(z)w^i \\&= \sum_{i=0}^\infty \sum_{n=i}^\infty \binom n i w^iz^n \\
&=\sum_{n=0}^\infty \sum_{i=0}^n \binom n i w^iz^n \\
&=\sum_{n=0}^\infty z^n(1+w)^n = \\
&=\frac{1}{1-z(1+w)}\\
&=\frac{1}{1-z}\frac{1}{1-w\frac{z}{1-z}}\\
&=\frac{1}{1-z}\sum_{i=0}^\infty \left(\frac{z}{1-z}\right)^iw^i
\end{align}$$
This means that $$S_i(z)=\frac{1}{1-z}\left(\frac z {1-z}\right)^i$$
The value you want is $zS_i(z) = \left(\frac z {1-z}\right)^{i+1}$, with $z=\frac 1 3$, $\frac z {1-z} = \frac 1 2$, so the answer should be $$\frac{1}{2^{i+1}}$$
A: Use the identity $${k\choose n}={k+1\choose n+1}-{k\choose n+1}$$ and let $$S_n = \sum_{k\geqslant n}{k\choose n}x^{k}$$ to get 
$$S_n=\sum_{k\geqslant n+1}{k\choose n}x^{k}+\color{Red}{x^n}=
\sum_{k\geqslant n+1}{k+1\choose n+1}x^k-\sum_{k\geqslant n+1}{k\choose n+1}x^k+\color{Red}{x^n}\\=
\color{Red}{x^{-1}}\left(\color{Blue}{\sum_{k\geqslant n+1}{k\choose n+1}x^k}-\color{Red}{x^{n+1}}\right)-\color{Blue}{\sum_{k\geqslant n+1}{k\choose n+1}x^k}+\color{Red}{x^n}\\=(x^{-1}-1)\color{Blue}{S_{n+1}}.$$ That way, $$\frac{S_n}{S_0}=\prod_{k=0}^{n-1}\frac{S_{k+1}}{S_k}=\left(\frac{1}{x^{-1}-1}\right)^n.$$ We know $S_0=\frac{1}{1-x}$, so $$S_n=\sum_{k\geqslant n}{k\choose n}x^k=x^{-1}\left(\frac{1}{x^{-1}-1}\right)^{n+1}.$$
