The value of $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}}$?
(a) 0
(b) $\dfrac{1}{3}$
(c) $\dfrac{1}{2}$
(d) 1
Answer
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Hint: Add the term ${{r}^{2}}$ in the denominator of the expression inside the summation and to balance it subtract the same. Now, use the algebraic identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$ to simplify the denominator. Further use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to write the denominator as a product of two terms. Now, write the numerator as the difference of these two terms of the product and break them into two fractions. Form a series and take the summation up to n terms, cancel the like terms and take the limit \[\displaystyle \lim_{n \to \infty }\] and substitute $n=\infty $ after converting the expression into the determinate form.
Complete step by step answer:
Here we have been provided with the expression $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}}$ and we are asked to find its value. Let us assume the expression $\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}$ as ${{K}_{r}}$. So we have,
$\Rightarrow {{K}_{r}}=\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}$
Adding and subtracting the term ${{r}^{2}}$ will have no effect on the expression, so we get,
$\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1+{{r}^{2}}-{{r}^{2}}} \\
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{4}}+2{{r}^{2}}+1 \right)-{{r}^{2}}} \\
\end{align}$
Using the algebraic identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$ we get,
$\Rightarrow {{K}_{r}}=\dfrac{r}{{{\left( {{r}^{2}}+1 \right)}^{2}}-{{r}^{2}}}$
Now, using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can write the expression as:
\[\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{2}}+1+r \right)\left( {{r}^{2}}+1-r \right)} \\
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \\
\end{align}\]
We can also simplify the numerator as $r=\dfrac{1}{2}\left[ \left( {{r}^{2}}+r+1 \right)-\left( {{r}^{2}}-r+1 \right) \right]$ so we get,
\[\Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{\left( {{r}^{2}}+r+1 \right)-\left( {{r}^{2}}-r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \right]\]
Breaking the terms we get,
\[\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{\left( {{r}^{2}}+r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)}-\dfrac{\left( {{r}^{2}}-r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \right] \\
& \Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{r}^{2}}+r+1 \right)} \right].......\left( i \right) \\
\end{align}\]
Now let us simplify the term \[\left( {{r}^{2}}+1+r \right)\]. So we can write,
\[\begin{align}
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{r}^{2}}+1+2r-r \\
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{\left( r+1 \right)}^{2}}-r-1+1 \\
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \\
\end{align}\]
Substituting the obtained expression in the denominator of the second term of equation (i) we get,
\[\Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]\]
Taking summation both the sides from r = 1 to r = n we get,
\[\begin{align}
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\sum\limits_{r=1}^{n}{\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]} \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]}.......\left( ii \right) \\
\end{align}\]
Now, let us see what happens if we substitute the values of r = 1, 2, 3,…., n in the expression ${{K}_{r}}$.
\[\begin{align}
& {{K}_{1}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{1}^{2}}-1+1 \right)}-\dfrac{1}{\left( {{\left( 2 \right)}^{2}}-\left( 2 \right)+1 \right)} \right] \\
& {{K}_{2}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{2}^{2}}-2+1 \right)}-\dfrac{1}{\left( {{\left( 3 \right)}^{2}}-\left( 3 \right)+1 \right)} \right] \\
& {{K}_{3}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{3}^{2}}-3+1 \right)}-\dfrac{1}{\left( {{\left( 4 \right)}^{2}}-\left( 4 \right)+1 \right)} \right] \\
\end{align}\]
This will carry up to r = n. What we see is if we add these n terms every second term of a expression will cancel every first term of its next expression, so on cancelling the like terms up to r = n we will have the expression (ii) left as:
\[\begin{align}
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{1}^{2}}-1+1 \right)}-\dfrac{1}{\left( {{\left( n+1 \right)}^{2}}-\left( n+1 \right)+1 \right)} \right] \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ 1-\dfrac{1}{\left( {{n}^{2}}+n+1 \right)} \right] \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{{{n}^{2}}+n}{\left( {{n}^{2}}+n+1 \right)} \right] \\
\end{align}\]
Taking ${{n}^{2}}$ common and cancelling we get,
\[\Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right]\]
Taking limit $\displaystyle \lim_{n \to \infty }$ both the sides we get,
\[\begin{align}
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\displaystyle \lim_{n \to \infty }\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right] \\
\end{align}\]
Substituting $n=\infty $ we get,
\[\begin{align}
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{\infty }}{\left( 1+\dfrac{1}{\infty }+\dfrac{1}{{{\infty }^{2}}} \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+0}{\left( 1+0+0 \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ 1 \right] \\
& \therefore \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2} \\
\end{align}\]
So, the correct answer is “Option c”.
Note: Here do not get confused by thinking that here we have to convert the limit into the definite integral because we cannot simplify the expression into the form $\dfrac{1}{n}f\left( \dfrac{r}{n} \right)$ which is necessary for the conversion. Sometimes you may get the series where you will be required to convert it into the inverse tangent function. In such cases you need to reduce the expression into the form $\left( \dfrac{a-b}{1+ab} \right)$ and then replace it with $\left( {{\tan }^{-1}}a-{{\tan }^{-1}}b \right)$.
Complete step by step answer:
Here we have been provided with the expression $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}}$ and we are asked to find its value. Let us assume the expression $\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}$ as ${{K}_{r}}$. So we have,
$\Rightarrow {{K}_{r}}=\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1}$
Adding and subtracting the term ${{r}^{2}}$ will have no effect on the expression, so we get,
$\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{r}{{{r}^{4}}+{{r}^{2}}+1+{{r}^{2}}-{{r}^{2}}} \\
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{4}}+2{{r}^{2}}+1 \right)-{{r}^{2}}} \\
\end{align}$
Using the algebraic identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$ we get,
$\Rightarrow {{K}_{r}}=\dfrac{r}{{{\left( {{r}^{2}}+1 \right)}^{2}}-{{r}^{2}}}$
Now, using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can write the expression as:
\[\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{2}}+1+r \right)\left( {{r}^{2}}+1-r \right)} \\
& \Rightarrow {{K}_{r}}=\dfrac{r}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \\
\end{align}\]
We can also simplify the numerator as $r=\dfrac{1}{2}\left[ \left( {{r}^{2}}+r+1 \right)-\left( {{r}^{2}}-r+1 \right) \right]$ so we get,
\[\Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{\left( {{r}^{2}}+r+1 \right)-\left( {{r}^{2}}-r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \right]\]
Breaking the terms we get,
\[\begin{align}
& \Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{\left( {{r}^{2}}+r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)}-\dfrac{\left( {{r}^{2}}-r+1 \right)}{\left( {{r}^{2}}+r+1 \right)\left( {{r}^{2}}-r+1 \right)} \right] \\
& \Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{r}^{2}}+r+1 \right)} \right].......\left( i \right) \\
\end{align}\]
Now let us simplify the term \[\left( {{r}^{2}}+1+r \right)\]. So we can write,
\[\begin{align}
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{r}^{2}}+1+2r-r \\
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{\left( r+1 \right)}^{2}}-r-1+1 \\
& \Rightarrow \left( {{r}^{2}}+1+r \right)={{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \\
\end{align}\]
Substituting the obtained expression in the denominator of the second term of equation (i) we get,
\[\Rightarrow {{K}_{r}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]\]
Taking summation both the sides from r = 1 to r = n we get,
\[\begin{align}
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\sum\limits_{r=1}^{n}{\dfrac{1}{2}\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]} \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\sum\limits_{r=1}^{n}{\left[ \dfrac{1}{\left( {{r}^{2}}-r+1 \right)}-\dfrac{1}{\left( {{\left( r+1 \right)}^{2}}-\left( r+1 \right)+1 \right)} \right]}.......\left( ii \right) \\
\end{align}\]
Now, let us see what happens if we substitute the values of r = 1, 2, 3,…., n in the expression ${{K}_{r}}$.
\[\begin{align}
& {{K}_{1}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{1}^{2}}-1+1 \right)}-\dfrac{1}{\left( {{\left( 2 \right)}^{2}}-\left( 2 \right)+1 \right)} \right] \\
& {{K}_{2}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{2}^{2}}-2+1 \right)}-\dfrac{1}{\left( {{\left( 3 \right)}^{2}}-\left( 3 \right)+1 \right)} \right] \\
& {{K}_{3}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{3}^{2}}-3+1 \right)}-\dfrac{1}{\left( {{\left( 4 \right)}^{2}}-\left( 4 \right)+1 \right)} \right] \\
\end{align}\]
This will carry up to r = n. What we see is if we add these n terms every second term of a expression will cancel every first term of its next expression, so on cancelling the like terms up to r = n we will have the expression (ii) left as:
\[\begin{align}
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1}{\left( {{1}^{2}}-1+1 \right)}-\dfrac{1}{\left( {{\left( n+1 \right)}^{2}}-\left( n+1 \right)+1 \right)} \right] \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ 1-\dfrac{1}{\left( {{n}^{2}}+n+1 \right)} \right] \\
& \Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{{{n}^{2}}+n}{\left( {{n}^{2}}+n+1 \right)} \right] \\
\end{align}\]
Taking ${{n}^{2}}$ common and cancelling we get,
\[\Rightarrow \sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right]\]
Taking limit $\displaystyle \lim_{n \to \infty }$ both the sides we get,
\[\begin{align}
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\displaystyle \lim_{n \to \infty }\left[ \dfrac{1+\dfrac{1}{n}}{\left( 1+\dfrac{1}{n}+\dfrac{1}{{{n}^{2}}} \right)} \right] \\
\end{align}\]
Substituting $n=\infty $ we get,
\[\begin{align}
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+\dfrac{1}{\infty }}{\left( 1+\dfrac{1}{\infty }+\dfrac{1}{{{\infty }^{2}}} \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ \dfrac{1+0}{\left( 1+0+0 \right)} \right] \\
& \Rightarrow \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2}\left[ 1 \right] \\
& \therefore \displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{{{K}_{r}}}=\dfrac{1}{2} \\
\end{align}\]
So, the correct answer is “Option c”.
Note: Here do not get confused by thinking that here we have to convert the limit into the definite integral because we cannot simplify the expression into the form $\dfrac{1}{n}f\left( \dfrac{r}{n} \right)$ which is necessary for the conversion. Sometimes you may get the series where you will be required to convert it into the inverse tangent function. In such cases you need to reduce the expression into the form $\left( \dfrac{a-b}{1+ab} \right)$ and then replace it with $\left( {{\tan }^{-1}}a-{{\tan }^{-1}}b \right)$.
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