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Question: Find the value of the following limit. \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1...

Find the value of the following limit.
lim⁡n→∞ ∑r=1ntan⁡−1(2r1−r2+r4)\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right)}
(a)π4\left( a \right)\dfrac{\pi }{4}
(b)π2\left( b \right)\dfrac{\pi }{2}
(c)3π4\left( c \right)\dfrac{3\pi }{4}
(d) None of these

Explanation

Solution

First we will write tan⁡−1(2r1−r2+r4){{\tan }^{-1}}\left( \dfrac{2r}{1-{{r}^{2}}+{{r}^{4}}} \right) in the form of tan⁡−1(a−b1+ab){{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right) and use the conversion formula tan⁡−1(a−b1+ab)=tan⁡−1a−tan⁡−1b.{{\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{{\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: lim⁡n→∞ ∑r=1ntan⁡−1(2r1−r2+r4).\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’.
⇒E=lim⁡n→∞ ∑r=1ntan⁡−1(2r1−r2+r4)\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
⇒E=lim⁡n→∞ ∑r=1ntan⁡−1(2r1+r2(r2−1))\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)}
⇒E=lim⁡n→∞ ∑r=1ntan⁡−1(2r1+r2(r−1)(r+1))\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)}
⇒E=lim⁡n→∞ ∑r=1ntan⁡−1(2r1+(r2−r)(r2+r))\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)}
⇒E=lim⁡n→∞ ∑r=1ntan⁡−1((r2+r)−(r2−r)1+(r2−r)(r2+r))\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(a−b1+ab){{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right) and we know that, tan⁡−1(a−b1+ab)=tan⁡−1a−tan⁡−1b.{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)={{\tan }^{-1}}a-{{\tan }^{-1}}b.
Therefore, the required expression becomes,
⇒E=lim⁡n→∞ ∑r=1n[tan⁡−1(r2+r)−tan⁡−1(r2−r)]\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]}
⇒E=lim⁡n→∞ ∑r=1n[tan⁡−1r(r+1)−tan⁡−1r(r−1)]\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,

& \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{aligned} \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.