Question

The value of \[{\tan ^{ - 1}}\dfrac{1}{3} + {\tan ^{ - 1}}\dfrac{1}{7} + {\tan ^{ - 1}}\dfrac{1}{{13}} + - - - + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + - - - \infty \] Is equal to ______

Answer 1

Hint: We can observe that in the given problem they are given the \[{n^{th}}\] term \[\dfrac{1}{{{n^2} + n + 1}}\] . By substituting the values of n=1, 2, 3… we get \[\dfrac{1}{3}\] , \[\dfrac{1}{7}\] \[\dfrac{1}{{13}}\] ----. In the \[{n^{th}}\] term the denominator as 1, adding and subtracting n and simplifying in the denominator we get \[{\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)\] . Using summation and formula we get the required solution.

Complete step-by-step answer:
Given, \[{\tan ^{ - 1}}\dfrac{1}{3} + {\tan ^{ - 1}}\dfrac{1}{7} + {\tan ^{ - 1}}\dfrac{1}{{13}} + - - - + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + - - - \infty \] .
Take \[{n^{th}}\] term as \[{\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right)\] .
Adding and subtracting n on the numerator.
 \[ = {\tan ^{ - 1}}\left( {\dfrac{{n + 1 - n}}{{{n^2} + n + 1}}} \right)\]
But the denominator term we can rearrange it as \[{n^2} + n + 1 = 1 + n + {n^2}\]
  \[ = 1 + n(1 + n)\]
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{n + 1 - n}}{{1 + n(1 + n)}}} \right)\] ----- (1)
We know the formula \[{\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right)\] , comparing this with equation (1).
 \[A = n + 1\] and \[B = n\]
 \[\therefore \] Equation (1) becomes \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)\] . ----- (2)
Now \[{S_\infty } = {\tan ^{ - 1}}\dfrac{1}{3} + {\tan ^{ - 1}}\dfrac{1}{7} + {\tan ^{ - 1}}\dfrac{1}{{13}} + - - - + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + - - - \infty \]
It can be written in a summation form,
 \[ = \sum\limits_{n = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right)} \]
If you put n values from n=1 to \[\infty \] . You will get the given problem.
Using equation (2) and substituting we get,
 \[ = \sum\limits_{n = 1}^\infty {[{{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}(n)] } \]
Put n=1 to \[\infty \] we get,
 \[ = \left( {{{\tan }^{ - 1}}(2) - {{\tan }^{ - 1}}(1)} \right) + \left( {{{\tan }^{ - 1}}(3) - {{\tan }^{ - 1}}(2)} \right) + \left( {{{\tan }^{ - 1}}(4) - {{\tan }^{ - 1}}(3)} \right) + - - - + \left( {{{\tan }^{ - 1}}(\infty ) - {{\tan }^{ - 1}}(\infty - 1)} \right) + \left( {{{\tan }^{ - 1}}(\infty + 1) - {{\tan }^{ - 1}}(\infty )} \right).\]
As we can see, \[{\tan ^{ - 1}}(2)\] will get canceled from first and second terms. \[{\tan ^{ - 1}}(3)\] will get cancels from the second term and third term. Similarly all the terms will get canceled. But in the last term \[{\tan ^{ - 1}}(\infty + 1)\] will remain as it is.
 \[ = - {\tan ^{ - 1}}(1) + - - - - - - + {\tan ^{ - 1}}(\infty + 1)\]
We know that \[{\tan ^{ - 1}}(1) = \dfrac{\pi }{4}\] and if we add some number to infinity we again get infinity \[{\tan ^{ - 1}}(\infty + 1) = {\tan ^{ - 1}}(\infty )\] . We know \[{\tan ^{ - 1}}(\infty ) = \dfrac{\pi }{2}\] substituting this we get,
 \[ = - {\tan ^{ - 1}}(1) + - - - - - - + {\tan ^{ - 1}}(\infty )\]
 \[ = - \dfrac{\pi }{4} + \dfrac{\pi }{2}\] (Taking L.C.M. and simplifying)
 \[ = \dfrac{\pi }{4}\]
Hence, \[{\tan ^{ - 1}}\dfrac{1}{3} + {\tan ^{ - 1}}\dfrac{1}{7} + {\tan ^{ - 1}}\dfrac{1}{{13}} + - - - + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + - - - \infty = \dfrac{\pi }{4}\] .
So, the correct answer is “$\dfrac{\pi }{4}$ .”.

Note: You can also solve this by changing the each numerator term of each term in the equation as we did in the nth term. It will be difficult to solve. So, we applied for the nth term and using summation we solved it. Careful with the cancellation after the removal of summation. Better write the last to terms I.e., \[n = \infty - 1\] and \[n = \infty \] (so that no terms will get missed).