Verified answer

EXPLANATION.

[tex]\sf \displaystyle S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + kn + k^{2} }[/tex]

[tex]\sf \displaystyle T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + kn + k^{2} }[/tex]

[tex]\sf \displaystyle n = 1, 2, 3. . . . .[/tex]

As we know that,

Definite integrals as a limits of sum.

[tex]\sf \displaystyle \int\limits^b_a {f(x)} \, dx = \lim_{h \to 0} h \sum_{r = 0}^{n - 1} f(a + rh)[/tex]

[tex]\sf \displaystyle where : b = a + nh[/tex]

[tex]\sf \displaystyle \lim_{n \to \infty} \frac{b - a}{n} \sum_{r = 0}^{n - 1} f \bigg( a + r . \frac{b - a}{n} \bigg)[/tex]

[tex]\sf \displaystyle let \ \ b = 1 \ \ and \ \ a = 0[/tex]

[tex]\sf \displaystyle \int\limits^1_0 {f(x)} \, dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} f \bigg(\frac{r}{n} \bigg)[/tex]

[tex]\sf \displaystyle \frac{r}{n} = x \ \ and \ \frac{1}{n} = dx[/tex]

[tex]\sf \displaystyle S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + kn + k^{2} }[/tex]

[tex]\sf \displaystyle S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} \bigg(1 + (k/n) + (k^{2}/n^{2} )\bigg)} < \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} \bigg[ \frac{1}{1 + (k/n) + (k/n)^{2} } \bigg][/tex]

[tex]\sf \displaystyle \int\limits^1_0\frac{1}{1 + x + x^{2} } dx[/tex]

[tex]\sf \displaystyle \int\limits^1_0 \frac{dx}{(x + 1/2)^{2} + (\sqrt{3} /2)^{2} }[/tex]

Formula of :

[tex]\sf \displaystyle \int \frac{dx}{x^{2} + a^{2} } = \frac{1}{a} tan^{-1} \bigg(\frac{x}{a} \bigg) + c[/tex]

Using this formula in the equation, we get.

[tex]\sf \displaystyle \bigg[\frac{1}{(\sqrt{3} /2)} tan^{-1} \bigg(\frac{(x + 1/2)}{(\sqrt{3} /2)} \bigg) \bigg]_{0}^{1}[/tex]

[tex]\sf \displaystyle \bigg[ \frac{2}{\sqrt{3} } tan^{-1} \bigg(\frac{2x + 1}{\sqrt{3} } \bigg) \bigg]_{0}^{1}[/tex]

As we know that,

In definite integration, first we put upper limits in the equation then we put lower limit in the equation.

[tex]\sf \displaystyle \frac{2}{\sqrt{3} } \bigg[ tan^{-1} \bigg(\frac{2(1) + 1}{\sqrt{3} } \bigg) - tan^{-1} \bigg(\frac{2(0) + 1}{\sqrt{3} } \bigg) \bigg][/tex]

[tex]\sf \displaystyle \frac{2}{\sqrt{3} } \bigg[tan^{-1} \sqrt{3} - tan^{-1} \bigg(\frac{1}{\sqrt{3} } \bigg) \bigg][/tex]

[tex]\sf \displaystyle \frac{2}{\sqrt{3} } \bigg[ \frac{\pi}{3} - \frac{\pi}{6} \bigg][/tex]

[tex]\sf \displaystyle \frac{2}{\sqrt{3} } \bigg[\frac{2 \pi - \pi}{6} \bigg][/tex]

[tex]\sf \displaystyle \frac{2}{\sqrt{3} } \bigg[\frac{\pi}{6} \bigg] = \frac{\pi}{3\sqrt{3} }[/tex]

[tex]\sf \displaystyle S_{n} < \frac{\pi}{3\sqrt{3} }[/tex]

Similarly,

[tex]\sf \displaystyle when \ \ T_{n} \ tends \ to \ \infty[/tex]

[tex]\sf \displaystyle T_{n} > \frac{\pi}{3\sqrt{3} }[/tex]

Option [A, D] is correct answer.

Your answer is in attachment

Hi

Good morning