Sum of infinite terms of the series `cot^(-1) ( 1^(2) + 3/4) + cot^(-1) ( 2^(2) + 3/4) + cot^(-1) ( 3^(2) + 3/4) + `... is

Options

A. ` pi//4`
B. `tan^(-1)2`
C. `tan^(-1) 3`
D. `tan^(-1) 4`

Answer

To find the sum of the infinite series given by

\[ S = \cot^{-1} \left(1^2 + \frac{3}{4}\right) + \cot^{-1} \left(2^2 + \frac{3}{4}\right) + \cot^{-1} \left(3^2 + \frac{3}{4}\right) + \ldots \]

we can express the \(n\)-th term of the series as:

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

### Step 1: Rewrite the cotangent inverse
Using the property of inverse trigonometric functions, we can rewrite \(T_n\):

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

### Step 2: Simplify the expression
We can further simplify the term:

\[ T_n = \tan^{-1} \left(\frac{1}{1 + n^2 - \frac{1}{4}}\right) = \tan^{-1} \left(\frac{1}{1 + n^2 - \left(\frac{1}{2}\right)^2}\right) \]

### Step 3: Apply the difference of squares
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we can express this as:

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

### Step 4: Use the tangent difference identity
We can express \(T_n\) in terms of the difference of two inverse tangents:

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

### Step 5: Write the series as a telescoping series
The sum \(S\) can be expressed as:

\[ S = \sum_{n=1}^{\infty} \left( \tan^{-1} \left(n + \frac{1}{2}\right) - \tan^{-1} \left(n - \frac{1}{2}\right) \right) \]

This is a telescoping series where most terms will cancel out.

### Step 6: Evaluate the limit
As \(n\) approaches infinity, the series simplifies to:

\[ S = \lim_{n \to \infty} \left( \tan^{-1} \left(n + \frac{1}{2}\right) - \tan^{-1} \left(\frac{1}{2}\right) \right) \]

Since \(\tan^{-1}(n + \frac{1}{2})\) approaches \(\frac{\pi}{2}\) as \(n\) approaches infinity, we have:

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

### Step 7: Use the cotangent identity
Using the identity \(\frac{\pi}{2} - \tan^{-1}(x) = \cot^{-1}(x)\), we can express the result as:

\[ S = \cot^{-1} \left(\frac{1}{2}\right) \]

### Step 8: Final result
Finally, we can express \(\cot^{-1} \left(\frac{1}{2}\right)\) as:

\[ S = \tan^{-1}(2) \]

Thus, the sum of the infinite series is:

\[ \boxed{\tan^{-1}(2)} \]