Summation #3

Calculus Level 3

∑ k = 1 ∞ 2 k k ! = e a − b \large \sum_{k=1}^\infty \frac{\sqrt{2^{k}}}{k!}=e^{\sqrt{\color{#D61F06} a}}-\color{#D61F06} b

What is a + b \color{#D61F06} a+b


The answer is 3.

Aidan Poor by Aidan Poor , 18, USA

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1 solution

Chew-Seong Cheong
Sep 26, 2018

Relevant wiki: Maclaurin Series

S = ∑ k = 1 ∞ 2 k k ! = ∑ k = 0 ∞ ( 2 ) k k ! − 1 By Maclaurin series e x = ∑ n = 0 ∞ x n n ! = e 2 − 1 \begin{aligned} S & = \sum_{\color{#3D99F6}k=1}^\infty \frac {\sqrt{2^k}}{k!} \\ & = {\color{#3D99F6}\sum_{\color{#D61F06}k=0}^\infty \frac {\left(\sqrt 2\right)^k}{k!}} - 1 & \small \color{#3D99F6} \text{By Maclaurin series }e^x = \sum_{n=0}^\infty \frac {x^n}{n!} \\ & = e^{\sqrt 2} - 1 \end{aligned}

Therefore, a + b = 2 + 1 = 3 a+b= 2+1 = \boxed 3 .

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