Question

\(\lim_{x\to\infty}\left(\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+\dfrac{n}{n^2+3^2}+\ldots+\dfrac{1}{5n}\right)\) is equal to :

• A. \(\tan^{-1}(2)\)
• B. \(\tan^{-1}(3)\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{2}\)

Hint:

The last term \(\frac{1}{5n}=\frac{n}{n^2+(2n)^2}\), so the sum is \(\sum_{r=1}^{2n}\frac{n}{n^2+r^2}\), a Riemann sum of \(\frac{1}{1+x^2}\) over [0, 2].

Answer 1

The Correct Option is A

The last term is \(\frac{1}{5n}=\frac{n}{n^2+(2n)^2}\), so the sum is \(\sum_{r=1}^{2n}\frac{n}{n^2+r^2}=\sum_{r=1}^{2n}\frac1n\cdot\frac{1}{1+(r/n)^2}\).
As \(n\to\infty\), with \(x=\frac rn\), \(dx=\frac1n\), r from 1 to 2n means x from 0 to 2.
Limit \(=\int_0^2\frac{dx}{1+x^2}=\left[\tan^{-1}x\right]_0^2=\tan^{-1}2\).
Answer: (1) \(\tan^{-1}(2)\)