The value of ` Sigma_(n=1)^(infty) cot^(-1) ( n^(2) + n +1)` is also equal to

Answer

To solve the problem of finding the value of the summation \( \sum_{n=1}^{\infty} \cot^{-1}(n^2 + n + 1) \), we will follow these steps:

### Step 1: Rewrite the cotangent inverse
We start with the expression:
\[ \sum_{n=1}^{\infty} \cot^{-1}(n^2 + n + 1) \]
Using the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \), we can rewrite the summation as:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right) \]

### Step 2: Simplify the argument of the tangent inverse
Next, we can factor the quadratic expression in the denominator:
\[ n^2 + n + 1 = n(n + 1) + 1 \]
This allows us to express the term as:
\[ \tan^{-1}\left(\frac{1}{n(n + 1) + 1}\right) \]

### Step 3: Use the tangent subtraction formula
We can use the formula for the difference of two tangent inverses:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab} \]
By setting \( a = n + 1 \) and \( b = n \), we can express:
\[ \tan^{-1}(n + 1) - \tan^{-1}(n) = \tan^{-1}\left(\frac{(n + 1) - n}{1 + n(n + 1)}\right) = \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right) \]

### Step 4: Rewrite the summation
Thus, we can rewrite our summation:
\[ \sum_{n=1}^{\infty} \left(\tan^{-1}(n + 1) - \tan^{-1}(n)\right) \]

### Step 5: Recognize the telescoping series
This summation is a telescoping series. When we expand it, most terms will cancel:
\[ (\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + (\tan^{-1}(4) - \tan^{-1}(3)) + \ldots \]
The only terms that do not get canceled are:
\[ \lim_{N \to \infty} \tan^{-1}(N + 1) - \tan^{-1}(1) \]

### Step 6: Evaluate the limit
As \( N \) approaches infinity, \( \tan^{-1}(N + 1) \) approaches \( \frac{\pi}{2} \):
\[ \lim_{N \to \infty} \tan^{-1}(N + 1) = \frac{\pi}{2} \]
Thus, we have:
\[ \frac{\pi}{2} - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Conclusion
Therefore, the value of the summation is:
\[ \sum_{n=1}^{\infty} \cot^{-1}(n^2 + n + 1) = \frac{\pi}{4} \]