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Question: The sum of the infinite series \( {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - ...

The sum of the infinite series cot⁡−12+cot⁡−18+cot⁡−118+cot⁡−132+……∞{\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + \ldots \ldots \infty is equal to
A. π4\dfrac{\pi }{4}
B. π3\dfrac{\pi }{3}
C. π6\dfrac{\pi }{6}
D. π8\dfrac{\pi }{8}

Explanation

Solution

Hint : Observe the series carefully and take it to the form of the summation series and replace the cotangent by tangent and try to make the tangent inverse identity for that take minor small steps in the numerator and the denominator.

Complete step-by-step answer :
The given statement is cot⁡−12+cot⁡−18+cot⁡−118+cot⁡−132+……∞{\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + \ldots \ldots \infty
On observing the numbers which are 2,8,18,32……∞2,8,18,32 \ldots \ldots \infty we come to know that the series if of the form of ∑2r2\sum {2{r^2}} here rr is a natural number.
So, replacing the given series with ∑r=1∞cot⁡−1(2r2)\sum\limits_{r = 1}^\infty {{{\cot }^{ - 1}}\left( {2{r^2}} \right)}
Using the property of trigonometric functions that the cotangent is inverse of tangent.
∑r=1∞cot⁡−1(2r2)=∑r=1∞tan⁡−1(12r2)\sum\limits_{r = 1}^\infty {{{\cot }^{ - 1}}\left( {2{r^2}} \right)} = \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{1}{{2{r^2}}}} \right)}
Now, I am doing some changes in the angle of tangent,
∑r=1∞tan⁡−1(12r2)\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{1}{{2{r^2}}}} \right)}
Multiply and divide the angle of tangent by 22
∑r=1∞tan⁡−1(24r2)\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{4{r^2}}}} \right)}
Now, adding and subtracting 11 in the denominator
∑r=1∞tan⁡−1(21+4r2−1)\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{1 + 4{r^2} - 1}}} \right)}
Now, forming the identity in the denominator a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
∑r=1∞tan⁡−1(21+(2r+1)(2r−1))\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{1 + \left( {2r + 1} \right)\left( {2r - 1} \right)}}} \right)}
Now, adding and subtracting 2r2r in the numerator and writing 2=1+12 = 1 + 1
So, ∑r=1∞tan⁡−1((2r+1)−(2r−1)1+(2r+1)(2r−1))\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{{\left( {2r + 1} \right) - \left( {2r - 1} \right)}}{{1 + \left( {2r + 1} \right)\left( {2r - 1} \right)}}} \right)}
All these steps come to form the identity of tangent that tan⁡−1(x−y1+xy)=tan⁡−1x−tan⁡−1y{\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y
Here x=2r+1,y=2r−1x = 2r + 1,y = 2r - 1
Hence, we are left with ∑r=1∞tan⁡−1(2r+1)−∑r=1∞tan⁡−1(2r−1)\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {2r + 1} \right) - \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {2r - 1} \right)} }
To solve the same taking the limits n→∞n \to \infty
lim⁡n→∞[∑r=1ntan⁡−1(2r+1)−∑r=1ntan⁡−1(2r−1)]\lim \mathop {}\nolimits_{n \to \infty } \left[ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {2r + 1} \right) - \sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {2r - 1} \right)} } } \right]
tan⁡−1∞−tan⁡−11 =π2−π4=π4   {\tan ^{ - 1}}\infty - {\tan ^{ - 1}}1 \\\ = \dfrac{\pi }{2} - \dfrac{\pi }{4} = \dfrac{\pi }{4} \;
So, the correct option is A.
So, the correct answer is “Option A”.

Note : In mathematics, firstly observe which formulas would take you one step closer to the solution and then proceed by taking the right steps. It Is not like that you have to learn the steps but we can form your own unique steps too.