Why $\sum_{i = j}^\infty\frac{i!}{(i-j)!\cdot j!}\cdot k^{i-j}\cdot (1-k)^j = 1/(1-k)$?

Question

$$ \sum_{i = j}^\infty\frac{i!}{(i-j)!\cdot j!}\cdot k^{i-j}\cdot (1-k)^j = \frac1{1-k} $$ I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.

Answer 1

it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $\sum_{l\ge 0}\binom{j+l}{j}k^l=(1-k)^{-j-1}$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=\binom{j+l}{j}(-1)^l$.