The value of sum `sum_(n=1)^(oo)cot^(-1)(((n^(2)+2n)(n^(2)+2n+1)+1)/(2n+2))` is equal to

Options

A. `cos^(-1)((1)/(sqrt(5)))`
B. `sec^(-1)((sqrt(5))/(2))`
C. `sin^(-1)((1)/(sqrt(5)))`
D. `cot^(-1)(1)`

Explanation

To solve the given summation problem, we need to evaluate the infinite series:

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

### Step 1: Simplify the Argument of Cotangent Inverse

First, we simplify the argument of the cotangent inverse:

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

Let \( x = n^2 + 2n \). Then, we can rewrite the expression as:

\[ \cot^{-1}\left(\frac{x(x + 1) + 1}{2(n + 1)}\right) \]

Calculating \( x(x + 1) + 1 \):

\[ x(x + 1) + 1 = n^4 + 2n^3 + n^2 + 1 \]

Thus, we have:

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

### Step 2: Rewrite the Expression

We can rewrite the cotangent inverse using properties of inverse trigonometric functions:

\[ \cot^{-1}(a) = \tan^{-1}\left(\frac{1}{a}\right) \]

Thus, we can express:

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

### Step 3: Analyze the Series

Next, we analyze the series:

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

### Step 4: Find the Limit

To find the limit of the series, we can use the properties of the inverse tangent function and its behavior as \( n \to \infty \):

\[ \lim_{n \to \infty} \tan^{-1}\left(\frac{2(n + 1)}{n^4 + 2n^3 + n^2 + 1}\right) \to 0 \]

### Step 5: Evaluate the Series

The series converges, and we can evaluate it using the telescoping nature of the terms:

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

This results in:

\[ \lim_{N \to \infty} \left(\tan^{-1}(N + 2) - \tan^{-1}(1)\right) = \frac{\pi}{2} - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Answer

Thus, the value of the sum is:

\[ \frac{\pi}{4} \]