The sum of the infinite series ^-1 ( 74) + ^-1 ( 194) + ^-1 ( 394) + ^-1 ( 674) + is:

The sum of the infinite series ^-1 ( 74) + ^-1 ( 194) + ^-1 ( 394… — Solution | JEE Mathematics

MCQMediumJEE 2025Sum of Series

The sum of the infinite series $\cot^{-1} \left( \frac{7}{4} \right) + \cot^{-1} \left( \frac{19}{4} \right) + \cot^{-1} \left( \frac{39}{4} \right) + \cot^{-1} \left( \frac{67}{4} \right) + \dots$ is:

A
$\frac{\pi}{2} + \tan^{-1} \left( \frac{1}{2} \right)$
B
$\frac{\pi}{2} - \cot^{-1} \left( \frac{1}{2} \right)$
C
$\frac{\pi}{2} + \cot^{-1} \left( \frac{1}{2} \right)$
D
$\frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right)$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The infinite series is

S = \cot^{-1} \left( \frac{7}{4} \right) + \cot^{-1} \left( \frac{19}{4} \right) + \cot^{-1} \left( \frac{39}{4} \right) + \cot^{-1} \left( \frac{67}{4} \right) + \dots

Find: Its sum.

Let the general term be $T_n$ . From the solution working,

T_n = \cot^{-1} \left( \frac{4n}{2n^2 + 3} \right)

This is rewritten as

T_n = \cot^{-1} \left( \frac{n + \frac{1}{2}}{1 + \left( n + \frac{1}{2} \right)^2} \right)

Telescoping Form from the Provided Working

Using the telescoping form stated in the solution,

S = T_1 + T_2 + \dots = \cot^{-1} \left( n + \frac{1}{2} \right) - \cot^{-1} \left( n - \frac{1}{2} \right)

Hence the series telescopes and the infinite sum is obtained as

S = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right)

Therefore, the correct option is D.

The second approach on the page states a different intermediate expression, but it also marks final answer as option 4. The authoritative conclusion on the page is

S = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right)

Common mistakes

Assuming the terms form a simple arithmetic pattern in the arguments and summing them directly is wrong, because inverse trigonometric series are usually handled through identities that create cancellation. Rewrite the terms in a telescoping-friendly form instead.
Using the identity for $\cot^{-1}(x) + \cot^{-1}(y)$ without checking branch values can lead to an incorrect constant such as $\pi$ instead of $\frac{\pi}{2}$ . Track the principal values carefully.
Confusing $\cot^{-1} \left( \frac{1}{2} \right)$ with $\tan^{-1} \left( \frac{1}{2} \right)$ is incorrect. Use the relation between inverse tangent and inverse cotangent carefully before matching the final option.

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Answer

Step-by-step solution

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