Find the sum of the series
`tan^(-1) 1/2+tan^(-1) 1/8+tan^(-1) 1/18+tan^(-1) 1/32+...oo`

Answer

To find the sum of the series

\[ S = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18} + \tan^{-1} \frac{1}{32} + \ldots \]

we first observe the pattern in the terms. The terms can be expressed as:

\[ A_i = \tan^{-1} \frac{1}{2i^2} \]

where \( i \) starts from 1 and goes to infinity. Therefore, we can rewrite the series as:

\[ S = \sum_{i=1}^{\infty} \tan^{-1} \frac{1}{2i^2} \]

### Step 1: Use the formula for the difference of arctangents

We can use the identity for the difference of arctangents:

\[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x - y}{1 + xy} \]

We want to express \( A_i \) in a form that allows us to use this identity. We can rewrite \( A_i \) as:

\[ A_i = \tan^{-1} \left( \frac{1}{2i^2} \right) = \tan^{-1} \left( \frac{1}{2(i^2)} \right) = \tan^{-1} \left( \frac{1}{2} \cdot \frac{1}{i^2} \right) \]

### Step 2: Express terms in a telescoping series

We can express \( A_i \) as:

\[ A_i = \tan^{-1} (2i - 1) - \tan^{-1} (2i + 1) \]

This means:

\[ S = \sum_{i=1}^{\infty} \left( \tan^{-1} (2i - 1) - \tan^{-1} (2i + 1) \right) \]

### Step 3: Recognize the telescoping nature

Notice that this sum is telescoping. When we write out the first few terms:

\[ S = \left( \tan^{-1} 1 - \tan^{-1} 3 \right) + \left( \tan^{-1} 3 - \tan^{-1} 5 \right) + \left( \tan^{-1} 5 - \tan^{-1} 7 \right) + \ldots \]

Most terms will cancel out, and we are left with:

\[ S = \lim_{n \to \infty} \left( \tan^{-1} (2n - 1) - \tan^{-1} 1 \right) \]

### Step 4: Evaluate the limit

As \( n \to \infty \), \( \tan^{-1} (2n - 1) \to \frac{\pi}{2} \). Therefore, we have:

\[ S = \frac{\pi}{2} - \tan^{-1} 1 \]

Since \( \tan^{-1} 1 = \frac{\pi}{4} \), we get:

\[ S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Result

Thus, the sum of the series is:

\[ \boxed{\frac{\pi}{4}} \]