Bon Voyage Summations!

First I will compute the sum $\sum_{k=1}^{\infty} (1-p)^{k-1} p$ . This is just a typical geometric series which evaluates to $1$ . Now I will find $\sum_{k=1}^{\infty} p (1-p)^{k-1} \sum_{n=0}^{k-1} \frac{e^{-t}t^n}{n!}$ .

Trying to compute that sum in the order presented seems to be impossible so I will change the order of the variables. We see that the ranges for $n$ and $k$ are $0 \leq n \leq k-1$ and $1 \leq k \leq \infty$ which is equivalent to $0 \leq n \leq \infty$ and $n+1 \leq k \leq \infty$ thus the sum is equal to: $\sum_{n=0}^{\infty} \frac{e^{-t} t^n}{n!} \sum_{k=n+1}^{\infty} p(1-p)^{k-1}$ . The inner sum is now a simple geometric series which evaluates to $(1-p)^n$ making our complete sum equal to $\sum_{n=0}^{\infty} \frac{e^{-t} t^n (1-p)^n}{n!}$ and we may unite both of the numbers being raised to the nth power to get $e^{-t} \sum_{n=0}^{\infty} \frac{(t - tp)^n}{n!}$ . That sum is now nothing more than the Taylor series for the exponential function so it evaluates to $e^{-t} e^{t - tp} = e^{-tp} = e^{-2016 \times \log 2} = (e^{\log 2})^{-2016} = 2^{-2016} = \left(\frac{1}{2} \right)^{2016}$ .

Uniting both results, we see that the sum of the problem is just equal to $1 - \left(\frac{1}{2} \right)^{2016}$ .