Finding Limits 5

Calculus Level 3

lim n ( n n 2 + 1 2 + n n 2 + 2 2 + n n 2 + 3 2 + + n n 2 + n 2 ) = ? \large \lim_{n\to\infty}\left(\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+\dfrac{n}{n^2+3^2}+\cdots+\dfrac{n}{n^2+n^2}\right) = \, ?

0 0 π 4 \frac{\pi}{4} π \pi \infty

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1 solution

Chew-Seong Cheong
Aug 17, 2016

Relevant wiki: Riemann Sums

L = lim n k = 1 n n n 2 + k 2 Divide up and down by n = lim n k = 1 n 1 n + k 2 n = lim n 1 n k = 1 n 1 1 + ( k n ) 2 By Riemann sums = a b 1 1 + x 2 d x a = lim n 1 n = 0 , b = lim n n n = 1 = 0 1 1 1 + x 2 d x = tan 1 x 0 1 = π 4 \begin{aligned} L & = \lim_{n \to \infty} \sum_{k=1}^n \frac n{n^2+k^2} & \small \color{#3D99F6}{\text{Divide up and down by }n} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac 1{n+\frac {k^2}n} \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \frac 1{1+\left(\frac kn\right)^2} & \small \color{#3D99F6}{\text{By Riemann sums}} \\ & = \int_a^b \frac 1{1+x^2} dx & \small \color{#3D99F6}{a = \lim_{n \to \infty} \frac 1n = 0, \ b = \lim_{n \to \infty} \frac nn = 1} \\ & = \int_0^1 \frac 1{1+x^2} dx \\ & = \tan^{-1} x \ \bigg|_0^1 \\ & = \boxed{\dfrac \pi 4} \end{aligned}

Thank you again for the elegant and clear solution :)

Hana Wehbi - 4 years, 10 months ago

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Nice to know that you like it. Thanks for the problem.

Chew-Seong Cheong - 4 years, 10 months ago

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