I have a sequence
$a_n = (1-p)^n \sum_{\frac{n}{2}\le k \le n} \binom{n}{k} \left( \frac{p}{1-p} \right)^k.$
I want to show that $a_n\to 0$ when $n\to\infty$ if $0\le p < \frac{1}{2}$. Here's a plot of the sequence for $p=\frac{1}{3}$:
Failed attempt:
$\begin{align} (1-p)^n \sum_{\frac{n}{2}\le k \le n} \binom{n}{k} \left( \frac{p}{1-p} \right)^k <& (1-p)^n \sum_{0 \le k \le n} \binom{n}{k}\left(\frac{p}{1-p}\right)^k \text{ for } n\ge 1\\ =& (1-p)^n \left(1+\frac{p}{1-p}\right)^n = 1. \end{align}$
My hope here was to end up with something like $0.99^n$ in that last step so I could argue that since $a_n<0.99^n$ and $0.99^n\to 0$ as $n\to\infty$, $a_n\to\infty$. Unfortunately it came out to 1, not $0.99^n$.