Limit + Summation = Integration (2)

Calculus Level 3

lim ⁡ n → ∞ ∑ k = 1 n e k n + e − k n n 11 − e 2 k n − e − 2 k n = sin ⁡ − 1 ( e a − e − a b ) \large \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{e^{\frac{k}{n}}+e^{-\frac{k}{n}}}{n\sqrt{11-e^{\frac{2k}{n}}-e^{-\frac{2k}{n}}}} = \sin^{-1}{\left(\frac{e^{a}-e^{-a}}{b}\right)}

If a a and b b above are positive integers, find a + b a+b .


The answer is 4.

Tommy Li by Tommy Li , 22, Hong Kong

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2 solutions

Mark Hennings
Jul 20, 2017

lim ⁡ n → ∞ ∑ k = 1 n e k n + e − k n n 11 − e 2 k n − e − 2 k n = ∫ 0 1 e x + e − x 11 − e 2 x − e − 2 x d x = ∫ 0 e − e − 1 d y 9 − y 2 = sin ⁡ − 1 ( e − e − 1 3 ) \lim_{n \to \infty} \sum_{k=1}^n \frac{e^{\frac{k}{n}} + e^{\frac{-k}{n}}}{n\sqrt{11 - e^{\frac{2k}{n}} - e^{-\frac{2k}{n}}}} \; = \; \int_0^1 \frac{e^x + e^{-x}}{\sqrt{11 - e^{2x} - e^{-2x}}}\,dx \; = \; \int_0^{e - e^{-1}} \frac{dy}{\sqrt{9 - y^2}} \; = \; \sin^{-1}\left(\frac{e - e^{-1}}{3}\right) using a Riemann sum and the substitution y = e x − e − x y = e^x - e^{-x} . The answer is 1 + 3 = 4 1 + 3 = \boxed{4} .

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Aaron Zhang
Jan 26, 2019

∫ 0 1 e x + e − x 9 − ( e x − e − x ) 2 d x \int_0^1{\frac{e^x+e^{-x}}{\sqrt {9-(e^x-e^{-x})^2}}dx}

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