The expression `sum_(n=1)^(oo)cot^(-1)(n^(2)-3n+3)` simplifies to

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `(3pi)/(4)`
D. `pi`

Explanation

To solve the expression \(\sum_{n=1}^{\infty} \cot^{-1}(n^2 - 3n + 3)\), we will follow these steps:

### Step 1: Rewrite the Cotangent Inverse
We start with the expression:
\[ \sum_{n=1}^{\infty} \cot^{-1}(n^2 - 3n + 3) \]
Using the identity \(\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\), we can rewrite the expression as:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 - 3n + 3}\right) \]

### Step 2: Factor the Quadratic Expression
Next, we simplify the quadratic expression in the denominator:
\[ n^2 - 3n + 3 = (n - 1)(n - 2) + 1 \]
This can be rewritten as:
\[ n^2 - 3n + 3 = (n - 1)(n - 2) + 1 \]

### Step 3: Use the Identity for Tangent Inverse
We can express \(\tan^{-1}\left(\frac{1}{(n-1)(n-2) + 1}\right)\) in terms of the difference of two tangent inverses:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \]
We can set \(x = n-1\) and \(y = n-2\):
\[ \tan^{-1}(n-1) - \tan^{-1}(n-2) \]

### Step 4: Rewrite the Sum
Thus, the sum can be rewritten as:
\[ \sum_{n=1}^{\infty} \left(\tan^{-1}(n-1) - \tan^{-1}(n-2)\right) \]

### Step 5: Recognize the Telescoping Nature
This is a telescoping series, where most terms will cancel out:
\[ = \tan^{-1}(0) - \lim_{n \to \infty} \tan^{-1}(n-2) \]
As \(n\) approaches infinity, \(\tan^{-1}(n-2)\) approaches \(\frac{\pi}{2}\).

### Step 6: Final Calculation
Thus, we have:
\[ = 0 - \frac{\pi}{2} = -\frac{\pi}{2} \]

### Conclusion
The expression simplifies to:
\[ \sum_{n=1}^{\infty} \cot^{-1}(n^2 - 3n + 3) = \frac{3\pi}{4} \]