Find the sum series: `tan^-1 (1/3)+tan^-1 (1/7)+tan^-1(1/13)+…to oo`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `(pi)/(3)`
D. `(pi)/(6)`

Answer

To find the sum of the series \( \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) + \tan^{-1} \left( \frac{1}{13} \right) + \ldots \) up to infinity, we will use the formula for the difference of inverse tangents:

\[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]

### Step-by-Step Solution:

1. **Identify the Series**:
We have the series \( S = \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) + \tan^{-1} \left( \frac{1}{13} \right) + \ldots \)

2. **Rewrite Terms**:
We can express the terms in the series in a form that allows us to apply the inverse tangent difference formula. Notice that the denominators can be expressed as \( 2n + 1 \) for \( n = 1, 3, 6, \ldots \) which corresponds to the sequence of odd numbers starting from 3.

3. **Use the Formula**:
We can write:
\[ S = \sum_{n=1}^{\infty} \tan^{-1} \left( \frac{1}{2n + 1} \right) \]

4. **Pair the Terms**:
Using the property of inverse tangent:
\[ \tan^{-1} \left( \frac{1}{2n + 1} \right) = \tan^{-1} \left( \frac{1}{2n - 1} \right) - \tan^{-1} \left( \frac{1}{2n + 1} \right) \]
This allows us to pair terms in the series.

5. **Cancellation**:
When we write out the series, we will notice that many terms cancel out:
\[ S = \left( \tan^{-1} \left( \frac{1}{1} \right) - \tan^{-1} \left( \frac{1}{3} \right) \right) + \left( \tan^{-1} \left( \frac{1}{3} \right) - \tan^{-1} \left( \frac{1}{5} \right) \right) + \ldots \]
This results in a telescoping series.

6. **Evaluate the Limit**:
As \( n \) approaches infinity, the last term approaches \( \tan^{-1} \left( \frac{1}{\infty} \right) = 0 \). Thus, we are left with:
\[ S = \lim_{n \to \infty} \left( \tan^{-1} \left( n + 1 \right) - \tan^{-1} \left( 1 \right) \right) \]
Since \( \tan^{-1} \left( n + 1 \right) \) approaches \( \frac{\pi}{2} \) as \( n \to \infty \), we have:
\[ S = \frac{\pi}{2} - \frac{\pi}{4} \]

7. **Final Result**:
Thus, the sum of the series is:
\[ S = \frac{\pi}{4} \]