Euler Sequence Space

Question

In the Euler sequence $\sum_{k=0}^{n}\binom{n}{k}(1-\alpha)^{n-k}\alpha^k=1$ by binomial law. However, I encountered another sum $$\sum_{n=k}^{\infty}\binom{n}{k}(1-\alpha)^{n-k}\alpha^k=\frac{1}{\alpha} $$which I am not getting. What does $\sum_{n=k}^{\infty}$ represents in the Euler matrix? Please help me with this sum.

Answer 1

Using the following generating function: $\sum\limits_{n=0}^\infty a^n\binom{n+k}{k}x^n=\frac{1}{{(1-ax)}^{k+1}}$

When $a=1$ and $x=1-\alpha$ and reindex the sum n start form k, we get:

$\sum\limits_{n=k}^\infty\binom{n}{k}(1-\alpha)^{n-k}=\frac{1}{{\alpha}^{k+1}}$

Substitute back to the original sum:

$\sum_\limits{n=k}^{\infty}\binom{n}{k}(1-\alpha)^{n-k}\alpha^k=\frac{\alpha^k}{\alpha^{k+1}}=\frac{1}{\alpha}$

You can find the "Ordinary Generating Functions" on this link: https://en.wikipedia.org/wiki/Generating_function#Bivariate_and_multivariate_generating_functions