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Mathematics Question on Integrals of Some Particular Functions

Let Sn=∑k=0nnn2+kn+k2 and Tn=∑k=0n−11n2+kn+k2, for S_n= \displaystyle \sum_{k=0}^n \frac{n}{n^2+kn+k^2} \, and \, T_n= \displaystyle \sum_{k=0}^{n-1} \frac{1}{n^2+kn+k^2} , \, for \, n=1,2,3n = 1 ,2 ,3 ,... Then,

A

Sn<π33S_n < \frac{\pi}{3\sqrt 3}

B

Sn>π33S_n > \frac{\pi}{3\sqrt 3}

C

Tn<π33T_n < \frac{\pi}{3\sqrt 3}

D

Tn>π33T_n > \frac{\pi}{3\sqrt 3}

Answer

Tn>π33T_n > \frac{\pi}{3\sqrt 3}

Explanation

Solution

Given, Sn = ∑k=0nnn2+kn+k2 \displaystyle \sum_{k=0}^n \frac{n}{n^2+kn+k^2}
=∑k=0n1n.(11+kn+k2n2)  <limn→ ∞ ∑k=0n1n.(11+kn+(kn)2)=\displaystyle \sum_{k=0}^n \frac{1}{n}.\Bigg(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\Bigg) \, \, < lim_{n \to \, \infty} \, \displaystyle \sum_{k=0}^n \frac{1}{n}.\Bigg(\frac{1}{1+\frac{k}{n}+\bigg(\frac{k}{n}\bigg)^2}\Bigg)
∫0111+x+x2dx=[23tan−1(23(x+12))]01\int_0^1 \frac{1}{1+x+x^2}dx = \bigg[ \frac{2}{\sqrt 3} tan^{-1} \bigg(\frac{2}{\sqrt 3}\bigg(x+\frac{1}{2}\bigg)\bigg)\bigg]_0^1
=23.(π3−π6)=π33=\frac{2}{\sqrt 3}.\bigg(\frac{\pi}{3}-\frac{\pi}{6}\bigg) =\frac{\pi}{3\sqrt 3}
i.e..,            Sn<π33i.e.., \, \, \, \, \, \, \, \, \, \, \, \, S_n < \frac{\pi}{3\sqrt 3}
Similarly,           Tn>π33 \, \, \, \, \, \, \, \, \, \, T_n > \frac{\pi}{3\sqrt 3}