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HMMT February

Algebra Difficulty 5.3 AIME, harder Find the answer

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

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let q=1−pq=1-p. Then ∑k=1∞(1−∑n=0k−1e−ttnn!)qk−1p=∑k=1∞qk−1p−∑k=1∞∑n=0k−1e−ttnn!qk−1p=1−∑k=1∞∑n=0k−1e−ttnn!qk−1p=1−∑n=0∞∑k=n+1∞e−ttnn!qk−1p=1−∑n=0∞e−ttnn!qn=1−∑n=0∞e−t(qt)nn!=1−e−teqt=1−e−pt \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−(12)20161-\left(\frac{1}{2}\right)^{2016}.

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