HMMT-Feb 2016 guts, Problem 23

Let $t=2016$ and $p=\ln 2$. Evaluate in closed form the sum \[ \sum_{k=1}^{\infty}\left(1-\sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !}\right)(1-p)^{k-1} p \]

Solution: Let $q=1-p$. Then \[ \begin{aligned} \sum_{k=1}^{\infty}\left(1-\sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !}\right) q^{k-1} p & =\sum_{k=1}^{\infty} q^{k-1} p-\sum_{k=1}^{\infty} \sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{k=1}^{\infty} \sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{n=0}^{\infty} \sum_{k=n+1}^{\infty} \frac{e^{-t} t^{n}}{n !} q^{k-1} p \\ & =1-\sum_{n=0}^{\infty} \frac{e^{-t} t^{n}}{n !} q^{n} \\ & =1-\sum_{n=0}^{\infty} \frac{e^{-t}(q t)^{n}}{n !}=1-e^{-t} e^{q t}=1-e^{-p t} \end{aligned} \] Thus the answer is $1-\left(\frac{1}{2}\right)^{2016}$. $\fbox{1-\left(\frac{1}{2}\right)^{2016}}$.

Answer: 1-\left(\frac{1}{2}\right)^{2016}