Identity for $\sum\limits_{j = a}^{N}     \binom{N}{j} \binom{j}{a} d^{-j}$? I have run across the following multinomial series:
$$     \sum_{j = a}^{N}
    \binom{N}{j} \binom{j}{a} d^{-j}
$$
Here, $d>1$. This seems like a formula which has either a well-known identity, or which has no further closed form or simplification. Can anyone shed any light on this formula, or possibly point me to a standard reference?

Suppose we seek to evaluate
$$\sum_{j=a}^N {N\choose j}{j\choose a} d^{-j}.$$
Start from
$${j\choose a}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{a+1}} (1+z)^j \; dz.$$
Note that with $j$ and $a$ integers when $j\lt a$ then $j-a \lt 0$ and
$j-(a+1) \lt  -1$ so we may extend  the sum to start  at $j=0$ because
the extra contribution is zero.
This gives for the sum the integral
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{a+1}} 
\sum_{j=0}^N {n\choose j} \times (1+z)^j \times d^{-j} \; dz.$$
This simplifies to
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{a+1}} 
\left(\frac{1+z}{d}+1\right)^N \; dz$$
which is
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{a+1}} 
\left(\frac{1+d+z}{d}\right)^N \; dz.$$
Hence the closed form is
$$[z^a] \left(\frac{1+d+z}{d}\right)^N
= \frac{1}{d^N} [z^a] (1+d+z)^N
= \frac{1}{d^N} {N\choose a} (1+d)^{N-a}$$
This finally yields
$$\frac{1}{d^a} \frac{1}{d^{N-a}} {N\choose a} (1+d)^{N-a}
= \frac{1}{d^a} {N\choose a} 
\left(1 + \frac{1}{d}\right)^{N-a}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
