Binomial series $\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$ Does someone know how to derive this sum?
$$\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$$ where $|a|<1$ and $k$ is given integer.
 A: Say $$S(k)=\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}\tag1$$
$$\Rightarrow aS(k)=\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n+1}=\sum^{\infty}_{n=1} {{n+k-1}\choose{k}} a^{n}\tag2$$
Hence, $$S(k)-aS(k)=\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}-\sum^{\infty}_{n=1} {{n+k-1}\choose{k}} a^{n}$$
$$\Rightarrow (1-a)S(k)=1+\sum^{\infty}_{n=1} {{n+k}\choose{k}} a^{n}-\sum^{\infty}_{n=1} {{n+k-1}\choose{k}} a^{n}$$
$$\Rightarrow (1-a)S(k)=1+ \sum^{\infty}_{n=1} \left(\frac{k}{n+k}\right){{n+k}\choose{k}} a^{n}$$
$$\Rightarrow (1-a)S(k)=1+ \sum^{\infty}_{n=1} {{n+k-1}\choose{k-1}} a^{n}$$
$$\Rightarrow (1-a)S(k)=S(k-1)$$
Hence, we get $\boxed{\frac{S(k)}{S(k-1)}=\frac{1}{1-a}}$ for $k=1,2,..$
So we take product of all ratios till $k=k$,
$$\frac{S(1)}{S(0)}\cdot \frac{S(2)}{S(1)}\cdot \frac{S(3)}{S(2)}\ldots\frac{S(k)}{S(k-1)}=\frac{1}{1-a} \cdot\frac{1}{1-a} \cdot\frac{1}{1-a} \ldots \frac{1}{1-a}$$
$$\frac{S(k)}{S(0)}=\left(\frac{1}{1-a}\right)^k$$
$S(0)$ is a simple geometric series equal to $\frac{1}{1-a}$.
Hence, the answer is $$\boxed{S(k)=\frac{1}{\left(1-a\right)^{k+1}}}$$
Hope this helps you.
A: Binomial series
We know by Taylor series 
$$(1-a)^{-\lambda} =  \sum^{\infty}_{n=0} \lambda(\lambda+1)\cdots(\lambda+n) \frac{a^{n}}{n!}= \sum^{\infty}_{n=0} \frac{\Gamma(n+\lambda)}{n!\Gamma{(\lambda)}} a^{n},\lambda>0.$$
but since $\Gamma(m+1)=m!$ for $m\in\mathbb N$
 $$\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n} =\sum^{\infty}_{n=0} \frac{(n+k)!}{n!k!} a^{n}=\frac{\Gamma(n+k+1)}{n!\Gamma{(k+1)}} a^{n} =(1-a)^{-k-1} $$
To see that, $$\lambda(\lambda+1)\cdots(\lambda+n) = \frac{\Gamma(n+\lambda)}{\Gamma{(\lambda)}}$$
Use $ \Gamma(1+\lambda) = \lambda\Gamma(\lambda).$
A: $\binom{n+k}{k} = [x^k]:(1+x)^{n+k}$
\begin{eqnarray*} 
\sum_{n=0}^{\infty} \binom{n+k}{k} a^k &=& [x^k]: \sum_{n=0}^{\infty}  a^k (1+x)^{n+k} \\
&=& [x^k]:    (1+x)^{k} \frac{1}{1-a(1+x)} \\
&=& [x^k]:    (1+x)^{k} \frac{1}{1-a} \frac{1}{1-\frac{ax}{1-a}}\\
&=& \frac{1}{1-a}[x^k]:   \left( \sum_{j=0}^{k} \binom{k}{j} x^{k-j} \right)\left( \sum_{l=0}^{\infty} \left(\frac{ax}{1-a} \right)^l \right) \\
&=&\frac{1}{1-a}\sum_{j=0}^{k} \binom{k}{j} \left(\frac{a}{1-a} \right)^j \\
&=&\frac{1}{1-a}\left(1+ \frac{a}{1-a}\right)^k = \color{blue}{\frac{1}{(1-a)^{k+1}}}
\end{eqnarray*}
