Summing the binomial pmf over $n$ I was trying to work out some bounds for a research problem when I came across the innocuous-looking sum:
$$ \sum_{n=k}^{\infty} {n \choose k} p^k (1-p)^{n-k}, \quad k \in \mathbb{N}, \; p \in (0,1)$$
At first I only needed to show that the sum converges, but later I realized that I need to actually do the sum, which looks familiar/easy but apparently isn't.
Trying various Binomial identities didn't go anywhere, but my repertoire is limited. I also eventually realized that for fixed $p$, the sums should not vary too much across $k$ since when $n$ gets big, the difference in $k$ don't matter much. But I'm not sure how to exploit this.

Let $q=1-p$; then you want
$$p^k\sum_{n\ge k}\binom{n}kq^{n-k}=p^k\sum_{\ell\ge 0}\binom{k+\ell}kq^\ell\;.$$
The generating function of the sequence $\left\langle\binom{k+\ell}k:\ell\in\Bbb N\right\rangle$ is
$$\sum_{\ell\ge 0}\binom{k+\ell}kx^\ell=\frac1{(1-x)^{k+1}}\;,$$
so
$$p^k\sum_{n\ge k}\binom{n}kq^{n-k}=\frac{p^k}{(1-q)^{k+1}}=\frac1p\;.$$
