# Let ${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}\text{ and }{{T}_{n}}=\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}$ , for $n=1,2,3,....$ Then,

(A) ${{S}_{n}}<\dfrac{\pi }{3\sqrt{3}}$

(B) ${{S}_{n}}>\dfrac{\pi }{3\sqrt{3}}$

(C) ${{T}_{n}}<\dfrac{\pi }{3\sqrt{3}}$

(D) ${{T}_{n}}>\dfrac{\pi }{3\sqrt{3}}$

Answer

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Hint: Don’t calculate exact${{T}_{n}}$and ${{S}_{n}}$. Just try to compare it with infinite summation of the given function.

We have the series given;

${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}\text{ }............\left( 1 \right)\text{ }$

${{T}_{n}}=\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}.............\left( 2 \right)$

It is tempting to start by evaluating both the sums in closed forms. But getting sums for the above series is usually difficult, but not impossible. But it is not necessary to calculate the exact sums. We need to compare ${{S}_{n}}$ and ${{T}_{n}}$by $\dfrac{\pi }{3\sqrt{3}}$ , from options.

The general terms of both the series are the same i.e., $\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}$.

If we recast it, we can divide by ${{n}^{2}}$ to numerator and denominator both, we get general term as;

\[\dfrac{\dfrac{1}{n}}{1+\left( \dfrac{k}{n} \right)+{{\left( \dfrac{k}{n} \right)}^{2}}}\]

Now, let us evaluate ${{S}_{n}}$ and ${{T}_{n}}$by following approach;

${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{n\left( 1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}} \right)}}\text{ }$

Let us compare the above summation with series;

$M=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{n}{n\left( 1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}} \right)}}$

Now, we can convert the above series of $\left( n\to \infty \right)$ to integral by using limit as a sum concept.

By Replacing

$\dfrac{1}{n}\text{ to }dx$

$\left( \dfrac{k}{n} \right)\text{ to }x$

Lower limit $=n\to \dfrac{k}{n}=0$

Upper limit $=$ maximum value of $k\text{ in }\dfrac{k}{n}=\dfrac{n}{n}=1$

Hence, we can rewrite series $M$ to definite integral as;

\[M=\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}}\]

Converting to square form, we get;

\[\begin{align}

& M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+1-\dfrac{1}{4}}} \\

& M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}} \\

\end{align}\]

We have;

$\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}$

Therefore, $M$can be written as;

\[M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}\]

\[M=\dfrac{1}{\left( \dfrac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left( \dfrac{\left( x+\dfrac{1}{2} \right)}{\left( \dfrac{\sqrt{3}}{2} \right)} \right)\left| \begin{matrix}

1 \\

0 \\

\end{matrix} \right.\]

Substituting the values, we get;

$\begin{align}

& M=\dfrac{2}{\sqrt{3}}\left( \dfrac{\pi }{3}-\dfrac{\pi }{6} \right) \\

& M=\dfrac{2}{\sqrt{3}}\dfrac{2\pi -\pi }{6}=\dfrac{2}{\sqrt{3}}.\dfrac{\pi }{6} \\

& M=\dfrac{\pi }{3\sqrt{3}}...............\left( 3 \right) \\

\end{align}$

Now, let us compare ${{T}_{n}},{{S}_{n}}\text{ and }M$.

We can observe that $\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}$is a decreasing function because it is converted to $\dfrac{1}{1+x+{{x}^{2}}}$where$\left( x=\dfrac{k}{n} \right)$ and $\dfrac{1}{1+x+{{x}^{2}}}$ or $\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}$is a decreasing function.

Thus, we see that$\left( \dfrac{\pi }{3\sqrt{3}} \right)$ is a magic number which is simply integral of function $\dfrac{1}{1+x+{{x}^{2}}}$ over the internal $\left( 0,1 \right)$. The integral is a decreasing function of $x$. So, when the internal $\left( 0,1 \right)$is divided to $n$ equal parts of length $\dfrac{1}{n}$, maximum and minimum of the function $f\left( x \right)$ over the ${{k}^{th}}$ sub-internal $\left[ \dfrac{k-1}{n},\dfrac{k}{n} \right]$ occurs at the left and right endpoints respectively.

As a result ${{S}_{n}}$ is lower sums of $f\left( x \right)$ for this partition.

Hence,

$\begin{align}

& {{T}_{n}}>\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}dx>{{S}_{n}}} \\

& {{T}_{n}}>\dfrac{\pi }{3\sqrt{3}}>{{S}_{n}} \\

\end{align}$

Hence, answer $\left( A,D \right)$ are correct.

Note: This question belongs to the Riemann sum of any function which is defined by limit as a sum of any function where we convert any given series to integral form.

One can waste his/her time for calculating exact sums of the given series which is a very complex approach and no need for that also.

Most confusing part is comparison between ${{T}_{n}},{{S}_{n}}\text{ and }\dfrac{\pi }{3\sqrt{3}}$which can be understood by dividing $\left( 0,1 \right)$to $n$ equal internals and determining series is increasing and decreasing as explained in solution.

We have the series given;

${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}\text{ }............\left( 1 \right)\text{ }$

${{T}_{n}}=\sum\limits_{k=0}^{n-1}{\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}}.............\left( 2 \right)$

It is tempting to start by evaluating both the sums in closed forms. But getting sums for the above series is usually difficult, but not impossible. But it is not necessary to calculate the exact sums. We need to compare ${{S}_{n}}$ and ${{T}_{n}}$by $\dfrac{\pi }{3\sqrt{3}}$ , from options.

The general terms of both the series are the same i.e., $\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}$.

If we recast it, we can divide by ${{n}^{2}}$ to numerator and denominator both, we get general term as;

\[\dfrac{\dfrac{1}{n}}{1+\left( \dfrac{k}{n} \right)+{{\left( \dfrac{k}{n} \right)}^{2}}}\]

Now, let us evaluate ${{S}_{n}}$ and ${{T}_{n}}$by following approach;

${{S}_{n}}=\sum\limits_{k=1}^{n}{\dfrac{n}{n\left( 1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}} \right)}}\text{ }$

Let us compare the above summation with series;

$M=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\dfrac{n}{n\left( 1+\dfrac{k}{n}+{{\left( \dfrac{k}{n} \right)}^{2}} \right)}}$

Now, we can convert the above series of $\left( n\to \infty \right)$ to integral by using limit as a sum concept.

By Replacing

$\dfrac{1}{n}\text{ to }dx$

$\left( \dfrac{k}{n} \right)\text{ to }x$

Lower limit $=n\to \dfrac{k}{n}=0$

Upper limit $=$ maximum value of $k\text{ in }\dfrac{k}{n}=\dfrac{n}{n}=1$

Hence, we can rewrite series $M$ to definite integral as;

\[M=\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}}\]

Converting to square form, we get;

\[\begin{align}

& M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+1-\dfrac{1}{4}}} \\

& M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}} \\

\end{align}\]

We have;

$\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}$

Therefore, $M$can be written as;

\[M=\int_{0}^{1}{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}\]

\[M=\dfrac{1}{\left( \dfrac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left( \dfrac{\left( x+\dfrac{1}{2} \right)}{\left( \dfrac{\sqrt{3}}{2} \right)} \right)\left| \begin{matrix}

1 \\

0 \\

\end{matrix} \right.\]

Substituting the values, we get;

$\begin{align}

& M=\dfrac{2}{\sqrt{3}}\left( \dfrac{\pi }{3}-\dfrac{\pi }{6} \right) \\

& M=\dfrac{2}{\sqrt{3}}\dfrac{2\pi -\pi }{6}=\dfrac{2}{\sqrt{3}}.\dfrac{\pi }{6} \\

& M=\dfrac{\pi }{3\sqrt{3}}...............\left( 3 \right) \\

\end{align}$

Now, let us compare ${{T}_{n}},{{S}_{n}}\text{ and }M$.

We can observe that $\dfrac{n}{{{n}^{2}}+kn+{{k}^{2}}}$is a decreasing function because it is converted to $\dfrac{1}{1+x+{{x}^{2}}}$where$\left( x=\dfrac{k}{n} \right)$ and $\dfrac{1}{1+x+{{x}^{2}}}$ or $\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}$is a decreasing function.

Thus, we see that$\left( \dfrac{\pi }{3\sqrt{3}} \right)$ is a magic number which is simply integral of function $\dfrac{1}{1+x+{{x}^{2}}}$ over the internal $\left( 0,1 \right)$. The integral is a decreasing function of $x$. So, when the internal $\left( 0,1 \right)$is divided to $n$ equal parts of length $\dfrac{1}{n}$, maximum and minimum of the function $f\left( x \right)$ over the ${{k}^{th}}$ sub-internal $\left[ \dfrac{k-1}{n},\dfrac{k}{n} \right]$ occurs at the left and right endpoints respectively.

As a result ${{S}_{n}}$ is lower sums of $f\left( x \right)$ for this partition.

Hence,

$\begin{align}

& {{T}_{n}}>\int_{0}^{1}{\dfrac{1}{1+x+{{x}^{2}}}dx>{{S}_{n}}} \\

& {{T}_{n}}>\dfrac{\pi }{3\sqrt{3}}>{{S}_{n}} \\

\end{align}$

Hence, answer $\left( A,D \right)$ are correct.

Note: This question belongs to the Riemann sum of any function which is defined by limit as a sum of any function where we convert any given series to integral form.

One can waste his/her time for calculating exact sums of the given series which is a very complex approach and no need for that also.

Most confusing part is comparison between ${{T}_{n}},{{S}_{n}}\text{ and }\dfrac{\pi }{3\sqrt{3}}$which can be understood by dividing $\left( 0,1 \right)$to $n$ equal internals and determining series is increasing and decreasing as explained in solution.

Last updated date: 01st Oct 2023

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Total views: 364.5k

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