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Question: The value of \[{\tan ^{ - 1}}\dfrac{1}{3} + {\tan ^{ - 1}}\dfrac{1}{7} + {\tan ^{ - 1}}\dfrac{1}{{13...

The value of tan⁡−113+tan⁡−117+tan⁡−1113+−−−+tan⁡−1(1n2+n+1)+−−−∞{\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 ______

Explanation

Solution

Hint : We can observe that in the given problem they are given the nth{n^{th}} term 1n2+n+1\dfrac{1}{{{n^2} + n + 1}} . By substituting the values of n=1, 2, 3… we get 13\dfrac{1}{3} , 17\dfrac{1}{7} 113\dfrac{1}{{13}} ----. In the nth{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){\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n) . Using summation and formula we get the required solution.

Complete step-by-step answer :
Given, tan⁡−113+tan⁡−117+tan⁡−1113+−−−+tan⁡−1(1n2+n+1)+−−−∞{\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 nth{n^{th}} term as tan⁡−1(1n2+n+1){\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) .
Adding and subtracting n on the numerator.
=tan⁡−1(n+1−nn2+n+1)= {\tan ^{ - 1}}\left( {\dfrac{{n + 1 - n}}{{{n^2} + n + 1}}} \right)
But the denominator term we can rearrange it as n2+n+1=1+n+n2{n^2} + n + 1 = 1 + n + {n^2}
=1+n(1+n)= 1 + n(1 + n)
⇒tan⁡−1(1n2+n+1)=tan⁡−1(n+1−n1+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⁡−1A−tan⁡−1B=tan⁡−1(A−B1+AB){\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right) , comparing this with equation (1).
A=n+1A = n + 1 and B=nB = n
∴\therefore Equation (1) becomes ⇒tan⁡−1(1n2+n+1)=tan⁡−1(n+1)−tan⁡−1(n) \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n) . ----- (2)
Now S∞=tan⁡−113+tan⁡−117+tan⁡−1113+−−−+tan⁡−1(1n2+n+1)+−−−∞{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,
=∑n=1∞tan⁡−1(1n2+n+1)= \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,
=∑n=1∞[tan⁡−1(n+1)−tan⁡−1(n)]= \sum\limits_{n = 1}^\infty {[{{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}(n)] }
Put n=1 to ∞\infty we get,
=(tan⁡−1(2)−tan⁡−1(1))+(tan⁡−1(3)−tan⁡−1(2))+(tan⁡−1(4)−tan⁡−1(3))+−−−+(tan⁡−1(∞)−tan⁡−1(∞−1))+(tan⁡−1(∞+1)−tan⁡−1(∞)).= \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){\tan ^{ - 1}}(2) will get canceled from first and second terms. tan⁡−1(3){\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(∞+1){\tan ^{ - 1}}(\infty + 1) will remain as it is.
=−tan⁡−1(1)+−−−−−−+tan⁡−1(∞+1)= - {\tan ^{ - 1}}(1) + - - - - - - + {\tan ^{ - 1}}(\infty + 1)
We know that tan⁡−1(1)=π4{\tan ^{ - 1}}(1) = \dfrac{\pi }{4} and if we add some number to infinity we again get infinity tan⁡−1(∞+1)=tan⁡−1(∞){\tan ^{ - 1}}(\infty + 1) = {\tan ^{ - 1}}(\infty ) . We know tan⁡−1(∞)=π2{\tan ^{ - 1}}(\infty ) = \dfrac{\pi }{2} substituting this we get,
=−tan⁡−1(1)+−−−−−−+tan⁡−1(∞)= - {\tan ^{ - 1}}(1) + - - - - - - + {\tan ^{ - 1}}(\infty )
=−π4+π2= - \dfrac{\pi }{4} + \dfrac{\pi }{2} (Taking L.C.M. and simplifying)
=π4= \dfrac{\pi }{4}
Hence, tan⁡−113+tan⁡−117+tan⁡−1113+−−−+tan⁡−1(1n2+n+1)+−−−∞=π4{\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 “π4\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=∞−1n = \infty - 1 and n=∞n = \infty (so that no terms will get missed).