A calculus problem by Jose Sacramento

Calculus Level 4

lim n ( n n 2 + 1 + n n 2 + 4 + n n 2 + 9 + + n n 2 + n 2 ) \lim_{n\to\infty} \left( \dfrac n{n^2 + 1} + \dfrac n{n^2+ 4} + \dfrac n{n^2 + 9} + \cdots + \dfrac n{n^2 + n^2} \right)

Find the value of the closed form of the above limit.

Give your answer to 3 decimal places.


The answer is 0.785.

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1 solution

Relevant wiki: Riemann Sums

L = lim n ( n n 2 + 1 + n n 2 + 4 + n n 2 + 9 + + n n 2 + n 2 ) = lim n k = 1 n n n 2 + k 2 Divide up and down with n 2 = lim n k = 1 n 1 n 1 + k 2 n 2 = lim n 1 n k = 1 n 1 1 + k 2 n 2 By Riemann sums = a = 0 b = 1 1 1 + x 2 d x a = lim n k n = 0 , b = lim n n n = 1 = tan 1 x 0 1 = π 4 0.785 \begin{aligned} L & = \lim_{n \to \infty} \left( \frac n{n^2+1} + \frac n{n^2+4} + \frac n{n^2+9} + \cdots + \frac n{n^2+n^2} \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac n{n^2+k^2} & \small \color{#3D99F6}{\text{Divide up and down with }n^2} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac {\frac 1n}{1+\frac {k^2}{n^2}} \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \frac 1{1+\frac {k^2}{n^2}} & \small \color{#3D99F6}{\text{By Riemann sums}} \\ & = \int_{a=0}^{b=1} \frac 1{1+x^2} dx & \small \color{#3D99F6}{a = \lim_{n \to \infty} \frac kn = 0, \ b = \lim_{n \to \infty} \frac nn = 1} \\ & = \tan^{-1} x \ \bigg|_0^1 = \frac \pi 4 \approx \boxed{0.785} \end{aligned}

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