# Series of Binomial Coefficients

• February 6th 2011, 01:31 AM
machi4velli
Series of Binomial Coefficients
I'm trying to solve an infinite series (it's part of a larger applied problem I'm working with, but I know what to do with this once I find it, so it's not relevant).

I need the sum from n = k to n = infinity of

C(n,k)*c^(-n)

for a positive integer c > 1.

It's a little different, sort of a reverse of the binomial theorem since the n runs with the series with k held constant.

I have found that

C(n+j,k) = C(n,k)*(n+j)!(n-k)!/[n!(n+j-k)!]

But that seemed to cause more harm than good...

I know from Mathematica the sum is going to be c/(c-1)^(k+1), but I'm having lots of trouble proving this.

Thanks in advance for any assistance.
• February 6th 2011, 02:23 AM
Unbeatable0
We want the coefficient of $y^k$ is the following (then put $x=c^{-1}$):

$\displaystyle{
\sum_{m=0}^\infty y^m\sum_{n=0}^\infty \binom{n}{m} x^n=\sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n}{m} y^mx^n
}$

$\displaystyle{
=\sum_{n=0}^\infty (1+y)^nx^n = \frac{1}{1-x-xy}=\frac{1}{1-x}\cdot\frac{1}{1-\frac{x}{1-x}y}
}$

$\displaystyle{
=\sum_{m=0}^\infty \frac{x^m}{(1-x)^{m+1}}y^m
}$

Therefore

$\displaystyle\sum_{n=0}^\infty \binom{n}{k}c^{-n}=\frac{c^{-k}}{(1-c^{-1})^{k+1}}=\frac{c}{(c-1)^{k+1}}$

Another way: let

$\displaystyle{
S_k=\sum_{n=0}^\infty x^n\binom{n}{k}
}$

Then, using $\binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k+1}$ which follows from Pascal's identity, we get

$\displaystyle{S_k=\left(\frac{1}{x}-1}\right)S_{k+1}}$

or equivalently:

$\displaystyle{
S_{k+1} = \frac{x}{1-x}S_k
}$

and therefore

$
\displaystyle{S_{k}=\frac{x^{k-1}}{(1-x)^{k-1}}\sum_{n=0}^\infty \binom{n}{1}x^n}=
\frac{x^{k-1}}{(1-x)^{k-1}}\sum_{n=0}^\infty nx^n}
$

$\displaystyle{=\frac{x^{k-1}}{(1-x)^{k-1}}\frac{x}{(1-x)^2}
$

As we got earlier.

Edit:
Actually, in the second method, it would be easier to go down to $S_0$ instead of $S_1$:

$
\displaystyle{S_{k}=\frac{x^k}{(1-x)^k}S_0 = \frac{x^k}{(1-x)^k}\cdot\frac{1}{1-x}
$