The value of `sum_(r=2)^(oo) tan^(-1)((1)/(r^(2)-5r+7))`, is

Options

A. `pi/4`
B. `pi/2`
C. `(3pi)/(4)`
D. `(5pi)/(4)`

Answer

To solve the problem \( \sum_{r=2}^{\infty} \tan^{-1}\left(\frac{1}{r^2 - 5r + 7}\right) \), we can follow these steps:

### Step 1: Simplify the Argument of the Arctangent
We start by rewriting the expression inside the summation:
\[ r^2 - 5r + 7 = (r^2 - 5r + 6) + 1 = (r-2)(r-3) + 1 \]
Thus, we can express the summation as:
\[ \sum_{r=2}^{\infty} \tan^{-1}\left(\frac{1}{(r-2)(r-3) + 1}\right) \]

### Step 2: Rewrite the Arctangent
Now, we can rewrite the term \( \tan^{-1}\left(\frac{1}{(r-2)(r-3) + 1}\right) \) using the identity:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right) \]
We can express \( \frac{1}{(r-2)(r-3) + 1} \) as:
\[ \tan^{-1}(r-2) - \tan^{-1}(r-3) \]
So we have:
\[ \sum_{r=2}^{\infty} \left( \tan^{-1}(r-2) - \tan^{-1}(r-3) \right) \]

### Step 3: Recognize the Telescoping Series
The series now takes the form of a telescoping series:
\[ \left( \tan^{-1}(0) - \tan^{-1}(-1) \right) + \left( \tan^{-1}(1) - \tan^{-1}(0) \right) + \left( \tan^{-1}(2) - \tan^{-1}(1) \right) + \ldots \]
Most terms cancel out, leaving us with:
\[ \lim_{n \to \infty} \left( \tan^{-1}(n-2) - \tan^{-1}(-1) \right) \]

### Step 4: Evaluate the Limits
As \( n \to \infty \), \( \tan^{-1}(n-2) \to \frac{\pi}{2} \) and \( \tan^{-1}(-1) = -\frac{\pi}{4} \). Thus, we have:
\[ \frac{\pi}{2} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \]

### Final Answer
Therefore, the value of the series is:
\[ \sum_{r=2}^{\infty} \tan^{-1}\left(\frac{1}{r^2 - 5r + 7}\right) = \frac{3\pi}{4} \]