Find the sum `cot^(-1) 2 + cot^(-1) 8 + cot^(-1) 18 + ...oo`

Explanation

To find the sum \( \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \ldots \) up to infinity, we can follow these steps:

### Step 1: Identify the Pattern
The terms can be rewritten in a more general form. Notice that the terms can be expressed as:
\[ \cot^{-1}(2n^2) \quad \text{for } n = 1, 2, 3, \ldots \]
This gives us the series:
\[ \sum_{n=1}^{\infty} \cot^{-1}(2n^2) \]

### Step 2: Use the Identity for Cotangent Inverse
We can use the identity:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Thus, we can rewrite our sum as:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2n^2}\right) \]

### Step 3: Apply the Addition Formula for Tangent Inverse
We can use the formula for the difference of two tangent inverses:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
We can express \( \tan^{-1}\left(\frac{1}{2n^2}\right) \) in terms of a telescoping series:
\[ \tan^{-1}(2n+1) - \tan^{-1}(2n-1) \]

### Step 4: Rewrite the Series
Thus, we can rewrite our sum:
\[ \sum_{n=1}^{\infty} \left( \tan^{-1}(2n+1) - \tan^{-1}(2n-1) \right) \]

### Step 5: Observe the Telescoping Nature
This series is telescoping, meaning that most terms will cancel out:
\[ \left( \tan^{-1}(3) - \tan^{-1}(1) \right) + \left( \tan^{-1}(5) - \tan^{-1}(3) \right) + \left( \tan^{-1}(7) - \tan^{-1}(5) \right) + \ldots \]
After cancellation, we are left with:
\[ \lim_{n \to \infty} \tan^{-1}(2n+1) - \tan^{-1}(1) \]

### Step 6: Evaluate the Limit
As \( n \to \infty \), \( \tan^{-1}(2n+1) \) approaches \( \frac{\pi}{2} \):
\[ \lim_{n \to \infty} \tan^{-1}(2n+1) = \frac{\pi}{2} \]
Thus, the sum becomes:
\[ \frac{\pi}{2} - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} \]

### Step 7: Final Result
Calculating this gives:
\[ \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
Therefore, the sum \( \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \ldots \) converges to:
\[ \boxed{\frac{\pi}{4}} \]