Question

Let

Answer 1

Answer:

$\fbox{A \: and \: C \: is \: correct \: answer}$

Step-by-step explanation:

Given data:

$\displaystyle\sf S = \sum_{k=1}^n \dfrac{n}{n^2+kn + k^2} and , \\ \displaystyle\sf T_n =\sum_{k=0}^{n-1}\dfrac{n}{n^2+kn+k^2} \\ \displaystyle\sf \: where \: n = 1,2,3,4....$

Solution:

$\displaystyle\sf S_n = \sum_{k=1}^n \dfrac{n}{n^2+kn + k^2} \\ \displaystyle\sf S_n < \lim_{n\to\infty}S_n \\ \displaystyle\sf \: Taking \: common \: \rightarrow \frac{n}{ {n}^{2} } \\ \displaystyle\sf S_n < \lim_{n\to\infty}\sum_{k=1}^n \dfrac{n}{n^2+kn + k^2} \\ \displaystyle\sf S_n < \lim_{n\to\infty}\sum_{k=1}^n \dfrac{1}{1+ \frac{k}{n} + (\frac{k}{n}) ^2} \cdot \frac{ \not n}{\not {n}^{2} }_n \\ \displaystyle\sf S_n < \frac{1}{n} \lim_{n\to\infty}\sum_{k=1}^n \dfrac{1}{1+ \frac{k}{n} + (\frac{k}{n}) ^2} \\ \displaystyle\sf$

$\: Assume \rightarrow \frac{k}{n} = x \\ \displaystyle\sf \: S_n < \frac{1}{n} \lim_{n\to\infty}\sum_{k=1}^n \dfrac{1}{1+ x + x ^2} \\ \displaystyle\sf Rewriting \rightarrow (1+x+ {x}^{2}) \: as \: {(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} \\ \displaystyle\sf \: S_n < \frac{1}{n} \lim_{n\to\infty}\sum_{k=1}^n \dfrac{1}{{(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2}} \\ \displaystyle\sf \frac{k}{n} \: = x \rightarrow \: \frac{1}{n} \: = dx \\ \displaystyle\sf \: S_n < \int_0^1 \dfrac{1}{{(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2}} \cdot \: dx$

We know that,

$\displaystyle\sf \: \int \: \frac{1}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} ( \frac{x}{a} ) + c$

Applying this formula to above step,

$\displaystyle\sf \: S_n < \large \frac{1}{ \frac{ \sqrt{3} }{2} } \cdot \: {tan}^{ - 1} ( \frac{ \frac{2x + 1}{\not2} }{ \frac{ \sqrt{3} }{\not2} } ) \\ \displaystyle\sf \: S_n < \large | \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{2x + 1}{ \sqrt{3} }})| _0^1$

$\displaystyle\sf \: S_n < \large \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{2 \times 1 + 1}{ \sqrt{3} }}) - {tan}^{ - 1} ( { \frac{2 \times 0 + 1}{ \sqrt{3} }}) \\ \displaystyle\sf \: S_n < \large \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{\sqrt{3}}{ 1 }}) - {tan}^{ - 1} ( { \frac{ 1}{ \sqrt{3} }}) \\ \displaystyle\sf \: S_n < \large \frac{2}{ { \sqrt{3} }} \cdot \: ( \frac{ \pi}{3} - \frac{ \pi}{6} ) \\ \displaystyle\sf \: S_n < \large \frac{\not2}{ { \sqrt{3} }} \cdot \: (\frac{ \pi}{\not6} ) \\ \\ \displaystyle\sf \: S_n < \large \frac{ \pi}{3 \sqrt{3} }$

We must be knowing that,

$\displaystyle\sf \:h \sum_{k=0}^{n - 1} \: \frac{k}{n} > \int_0^1 \: f(x)dx \: > h \sum_{k=1}^n \: \frac{k}{n}$

Similarly we can solve for Tn,

$\displaystyle\sf T_n =\sum_{k=0}^{n-1}\dfrac{n}{n^2+kn+k^2} \\ \\ \displaystyle\sf T_n > \lim_{n\to\infty}T_n \\ \displaystyle\sf T_n > \lim_{n\to\infty}\sum_{k=0}^{n - 1} \dfrac{n}{n^2+kn + k^2} \\ \displaystyle\sf \: Taking \: common \: \rightarrow \frac{n}{ {n}^{2} } \\ \displaystyle\sf T_n > \lim_{n\to\infty}\sum_{k=0}^{n - 1} \dfrac{1}{1+ \frac{k}{n} + (\frac{k}{n}) ^2} \cdot \frac{ \not n}{\not {n}^{2} }_n \\ \displaystyle\sf T_n > \frac{1}{n} \lim_{n\to\infty}\sum_{k=0}^{n - 1} \dfrac{1}{1+ \frac{k}{n} + (\frac{k}{n}) ^2}$

Assume K/n = x,

$\displaystyle\sf \: T_n > \frac{1}{n} \lim_{n\to\infty}\sum_{k=0}^{n - 1} \dfrac{1}{1+ x + x ^2} \\ \displaystyle\sf Rewriting \rightarrow (1+x+ {x}^{2}) \: as \: {(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} \\ \displaystyle\sf \: T_n > \frac{1}{n} \lim_{n\to\infty}\sum_{k=0}^{n - 1} \dfrac{1}{{(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2}} \\\displaystyle\sf \frac{k}{n} \: = x \rightarrow \: \frac{1}{n} \: = dx \\ \displaystyle\sf \: T_n > \int_0^1 \dfrac{1}{{(x + \frac{1}{2})}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2}} \cdot \: dx$

Now we know that,

$\displaystyle\sf \: \int \: \frac{1}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} ( \frac{x}{a} ) + c \\ \displaystyle\sf applying \: this \: formula \: to \: above \: step$

$\\ \displaystyle\sf \: T_n > \large \frac{1}{ \frac{ \sqrt{3} }{2} } \cdot \: {tan}^{ - 1} ( \frac{ \frac{2x + 1}{\not2} }{ \frac{ \sqrt{3} }{\not2} } ) \\ \displaystyle\sf \: T_n > \large | \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{2x + 1}{ \sqrt{3} }})| _0^1 \\ \displaystyle\sf \: T_n > \large \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{2 \times 1 + 1}{ \sqrt{3} }}) - {tan}^{ - 1} ( { \frac{2 \times 0 + 1}{ \sqrt{3} }}) \\ \displaystyle\sf \: T_n > \large \frac{2}{ { \sqrt{3} }} \cdot \: {tan}^{ - 1} ( { \frac{\sqrt{3}}{ 1 }}) - {tan}^{ - 1} ( { \frac{ 1}{ \sqrt{3} }}) \\ \displaystyle\sf \: T_n > \large \frac{2}{ { \sqrt{3} }} \cdot \: ( \frac{ \pi}{3} - \frac{ \pi}{6} ) \\ \displaystyle\sf \: T_n > \large \frac{\not2}{ { \sqrt{3} }} \cdot \: (\frac{ \pi}{\not6} ) \\ \\ \displaystyle\sf \: T_n > \large \frac{ \pi}{3 \sqrt{3} }$

This is the final answer!

Thanks for joining brainly community!