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Calculus Level 3

lim n k = 1 n 1 k 2 e x 2 d x \large \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} \int_{-\infty}^{\infty} e^{-x^2}\, dx

Evaluate the limit above. Give your answer up to the fourth decimal place.


The answer is 2.9155.

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1 solution

Chew-Seong Cheong
Oct 24, 2018

Relevant wiki: Riemann Zeta Function

L = lim n k = 1 n 1 k 2 e x 2 d x Since the integrand is even. = k = 1 1 k 2 2 0 e x 2 d x Error function erf ( z ) = 2 π 0 z e t 2 d t = ζ ( 2 ) π erf ( ) Riemann zeta function ζ ( s ) = n = 1 1 n s = π 2 6 π 1 2.9156 \begin{aligned} L & = {\color{#3D99F6} \lim_{n \to \infty} \sum_{k=1}^n \frac 1{k^2}} \color{#D61F06} \int_{-\infty}^\infty e^{-x^2} dx & \small \color{#D61F06} \text{Since the integrand is even.} \\ & = {\color{#3D99F6} \sum_{k=1}^\infty \frac 1{k^2}} \cdot \color{#D61F06} 2 \int_0^\infty e^{-x^2} dx & \small \color{#D61F06} \text{Error function erf }(z) = \frac 2{\sqrt \pi} \int_0^z e^{-t^2} \ dt \\ & = {\color{#3D99F6} \zeta(2)} \cdot \color{#D61F06} \sqrt \pi \ \text{erf }(\infty) & \small \color{#3D99F6} \text{Riemann zeta function }\zeta (s) = \sum_{n=1}^\infty \frac 1{n^s} \\ & = {\color{#3D99F6} \frac {\pi^2}6} \cdot \color{#D61F06} \sqrt \pi \cdot 1 \\ & \approx \boxed{2.9156} \end{aligned}

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