Question

Find the value of the following limit.
\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right)}\]
\[\left( a \right)\dfrac{\pi }{4}\]
\[\left( b \right)\dfrac{\pi }{2}\]
\[\left( c \right)\dfrac{3\pi }{4}\]
(d) None of these

Answer 1

Hint: First we will write \[{{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right)\] in the form of \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)\] and use the conversion formula \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)={{\tan }^{-1}}a-{{\tan }^{-1}}b.\] Then we will remove the summation sign and write the terms of \[{{\tan }^{-1}}\] functions. Then we will cancel the like terms. Now, we will evaluate the limit to get the answer.

Complete step-by-step solution:
We have been given the expression: \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right)}.\] Let us assume the value of this expression as ‘E’.
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right)}\]
This can be written as
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1+{{r}^{2}}\left( {{r}^{2}}-1 \right)} \right)}\]
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1+{{r}^{2}}\left( r-1 \right)\left( r+1 \right)} \right)}\]
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1+\left( {{r}^{2}}-r \right)\left( {{r}^{2}}+r \right)} \right)}\]
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{\left( {{r}^{2}}+r \right)-\left( {{r}^{2}}-r \right)}{1+\left( {{r}^{2}}-r \right)\left( {{r}^{2}}+r \right)} \right)}\]
The above expression is of the form \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)\] and we know that, \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)={{\tan }^{-1}}a-{{\tan }^{-1}}b.\]
Therefore, the required expression becomes,
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\left[ {{\tan }^{-1}}\left( {{r}^{2}}+r \right)-{{\tan }^{-1}}\left( {{r}^{2}}-r \right) \right]}\]
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\left[ {{\tan }^{-1}}r\left( r+1 \right)-{{\tan }^{-1}}r\left( r-1 \right) \right]}\]
Removing the summation sign, we have,
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \begin{align}
  & \left( {{\tan }^{-1}}1\times 2-{{\tan }^{-1}}1\times 0 \right)+\left( {{\tan }^{-1}}2\times 3-{{\tan }^{-1}}1\times 2 \right)+ \\
 & \left( {{\tan }^{-1}}3\times 4-{{\tan }^{-1}}2\times 3 \right)+.....+\left( {{\tan }^{-1}}n\left( n+1 \right)-{{\tan }^{-1}}n\left( n-1 \right) \right) \\
\end{align} \right]\]
Cancelling the like terms, we get,
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\left[ {{\tan }^{-1}}n\left( n+1 \right)-{{\tan }^{-1}}0 \right]\]
\[\Rightarrow E=\underset{n\to \infty }{\mathop{\lim }}\,\left[ {{\tan }^{-1}}n\left( n+1 \right) \right]\]
Substituting the value of \[n=\infty ,\] we get,
\[\Rightarrow E={{\tan }^{-1}}\infty \]
Since, \[{{\tan }^{-1}}\infty =\dfrac{\pi }{2},\] we can write
\[\Rightarrow E=\dfrac{\pi }{2}\]
Hence, the value of the given expression is \[\dfrac{\pi }{2}.\]
Therefore, option (b) is the right answer.

Note: One may note that we must not try to convert the tan inverse function in the form of \[{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right).\] This is because its formula is given by \[{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)={{\tan }^{-1}}a+{{\tan }^{-1}}b.\] If we use this conversion then we will not be able to cancel the like terms and finally we will not be able to evaluate the limit.