The sum of the infinite series cot^(-1)(7/4)+cot^(-1)(19/4)+cot^(-1)(39/4)+cot^(-1)(67/4)+... is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2025

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Home Maths Inverse TrigonometryQuestion #233

The sum of the infinite series

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

is:

Clue 1 of 3: Identify the General Term

Determine the pattern of the numerators in the arguments of the $\cot^{- 1}$ function to find the general term of the series.

🥷

Ninja Strategy•Magnitude and Identity Analysis

First, observe that the sum of the series must be less than $\frac{π}{2}$ to eliminate options that add a positive term to $\frac{π}{2}$ . Then, use the complementary identity $\cot^{- 1} x = \frac{π}{2} - \tan^{- 1} x$ to evaluate the remaining options and find the correct form.

🧠 Full Solution Path

Step 1: Determine the General Term of the Series✦ Active

Let the given series be $S = \sum_{n = 1}^{\infty} T_{n}$ . The arguments of $\cot^{- 1}$ are $\frac{7}{4}, \frac{19}{4}, \frac{39}{4}, \frac{67}{4}, \dots$ . Let the numerators be $a_{n} = 7, 19, 39, 67, \dots$ . The first differences are $12, 20, 28, \dots$ . The second differences are $8, 8, \dots$ . Since the second differences are constant, $a_{n}$ is a quadratic in $n$ . Let $a_{n} = A n^{2} + B n + C$ . Using the first few terms:

a_{1} = A + B + C = 7

Difference a_{2} - a_{1} = 3 A + B = 12

Difference a_{3} - a_{2} = 5 A + B = 20

Subtracting these: (5 A + B) - (3 A + B) = 2 A = 8 ⟹ A = 4

Substituting A = 4 into 3 A + B = 12 ⟹ 12 + B = 12 ⟹ B = 0

Substituting A = 4, B = 0 into A + B + C = 7 ⟹ 4 + 0 + C = 7 ⟹ C = 3

Thus, the general numerator is a_{n} = 4 n^{2} + 3.

The general term of the series is T_{n} = \cot^{- 1} (\frac{4 n^{2} + 3}{4}) = \cot^{- 1} (n^{2} + \frac{3}{4}) .

Step 2: Transform the General Term into a Telescoping Form○ Expand

Use the identity $\cot^{- 1} x = \tan^{- 1} (\frac{1}{x})$ to convert the term to $\tan^{- 1}$ :

T_{n} = \tan^{- 1} (\frac{1}{n^{2} + \frac{3}{4}}) = \tan^{- 1} (\frac{4}{4 n^{2} + 3}) .

We aim to express $T_{n}$ in the form $\tan^{- 1} (f (n)) - \tan^{- 1} (f (n - 1))$ using the identity $\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})$ . Let $x = \frac{2 n + 1}{2}$ and $y = \frac{2 n - 1}{2}$ . Then:

\tan^{- 1} (\frac{2 n + 1}{2}) - \tan^{- 1} (\frac{2 n - 1}{2}) = \tan^{- 1} (\frac{\frac{2 n + 1}{2} - \frac{2 n - 1}{2}}{1 + (\frac{2 n + 1}{2}) (\frac{2 n - 1}{2})})

= \tan^{- 1} (\frac{\frac{2}{2}}{1 + \frac{4 n^{2} - 1}{4}}) = \tan^{- 1} (\frac{1}{\frac{4 + 4 n^{2} - 1}{4}}) = \tan^{- 1} (\frac{4}{4 n^{2} + 3}) .

This matches $T_{n}$ . So, $T_{n} = \tan^{- 1} (\frac{2 n + 1}{2}) - \tan^{- 1} (\frac{2 n - 1}{2})$ .

💡 Teacher's Secret Hint

Remember to look for ways to factor the denominator to match the $1 + x y$ form, and the numerator to match $x - y$ .

Step 3: Calculate the Sum of the Infinite Series○ Expand

The series is a telescoping sum. Let $S_{N}$ be the sum of the first $N$ terms:

S_{N} = \sum_{n = 1}^{N} [\tan^{- 1} (\frac{2 n + 1}{2}) - \tan^{- 1} (\frac{2 n - 1}{2})]

= (\tan^{- 1} (\frac{3}{2}) - \tan^{- 1} (\frac{1}{2})) + (\tan^{- 1} (\frac{5}{2}) - \tan^{- 1} (\frac{3}{2})) + \dots + (\tan^{- 1} (\frac{2 N + 1}{2}) - \tan^{- 1} (\frac{2 N - 1}{2}))

= \tan^{- 1} (\frac{2 N + 1}{2}) - \tan^{- 1} (\frac{1}{2}) .

Now, take the limit as $N \to \infty$ to find the sum of the infinite series:

S = lim_{N \to \infty} S_{N} = lim_{N \to \infty} [\tan^{- 1} (\frac{2 N + 1}{2}) - \tan^{- 1} (\frac{1}{2})]

= \tan^{- 1} (\infty) - \tan^{- 1} (\frac{1}{2}) = \frac{π}{2} - \tan^{- 1} (\frac{1}{2}) .

This matches option 3.

💡 Teacher's Secret Hint

Ensure you correctly identify the terms that cancel out in a telescoping sum and evaluate the limit carefully.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Identify the General Term

🧠 Full Solution Path