Question

The sum of the infinite series $ {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + \ldots \ldots \infty $ is equal to
A. $ \dfrac{\pi }{4} $
B. $ \dfrac{\pi }{3} $
C. $ \dfrac{\pi }{6} $
D. $ \dfrac{\pi }{8} $

Answer 1

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 ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + \ldots \ldots \infty $
On observing the numbers which are $ 2,8,18,32 \ldots \ldots \infty $ we come to know that the series if of the form of $ \sum {2{r^2}} $ here $ r $ is a natural number.
So, replacing the given series with $ \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.
 $ \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,
 $ \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{1}{{2{r^2}}}} \right)} $
Multiply and divide the angle of tangent by $ 2 $
 $ \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{4{r^2}}}} \right)} $
Now, adding and subtracting $ 1 $ in the denominator
 $ \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{1 + 4{r^2} - 1}}} \right)} $
Now, forming the identity in the denominator $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $
 $ \sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{2}{{1 + \left( {2r + 1} \right)\left( {2r - 1} \right)}}} \right)} $
Now, adding and subtracting $ 2r $ in the numerator and writing $ 2 = 1 + 1 $
So, $ \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}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y $
Here $ x = 2r + 1,y = 2r - 1 $
Hence, we are left with $ \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 \to \infty $
 $ \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}}\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.