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

Explanation

To find the sum \( S = \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \ldots \) up to infinity, we first need to identify the pattern in the terms.

1. **Identify the general term**:
The terms appear to follow a pattern. The \( n^{th} \) term can be expressed as:
\[ a_n = \cot^{-1}(n^2 + 1) \]
where \( n = 1, 2, 3, \ldots \)

2. **Use the cotangent addition formula**:
We can use the identity:
\[ \cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1}\left(\frac{xy - 1}{x + y}\right) \]
to combine terms.

3. **Combine terms**:
Let's combine the first two terms:
\[ \cot^{-1}(2) + \cot^{-1}(8) = \cot^{-1}\left(\frac{2 \cdot 8 - 1}{2 + 8}\right) = \cot^{-1}\left(\frac{16 - 1}{10}\right) = \cot^{-1}\left(\frac{15}{10}\right) = \cot^{-1}\left(\frac{3}{2}\right) \]

4. **Continue combining terms**:
Next, we can combine \( \cot^{-1}\left(\frac{3}{2}\right) + \cot^{-1}(18) \):
\[ \cot^{-1}\left(\frac{3}{2}\right) + \cot^{-1}(18) = \cot^{-1}\left(\frac{\frac{3}{2} \cdot 18 - 1}{\frac{3}{2} + 18}\right) = \cot^{-1}\left(\frac{27 - 1}{\frac{3}{2} + 18}\right) = \cot^{-1}\left(\frac{26}{\frac{39}{2}}\right) = \cot^{-1}\left(\frac{52}{39}\right) = \cot^{-1}\left(\frac{4}{3}\right) \]

5. **Generalize the pattern**:
Continuing this process, we can see that the sum converges to a limit as \( n \) approaches infinity. The series can be simplified using the formula:
\[ S = \sum_{n=1}^{\infty} \cot^{-1}(n^2 + 1) \]
This series converges to \( \frac{\pi}{2} \).

6. **Final result**:
Therefore, the sum \( S \) can be expressed as:
\[ S = \frac{\pi}{2} \]