The value of `tan^(-1).(4)/(7) + tan^(-1).(4)/(19) + tan^(-1).(4)/(39) + tan^(-1).(4)/(67) ... oo` equals

Options

A. `tan^(-1) 1 + tan^(-1).(1)/(2) + tan^(-1).(1)/(3)`
B. `tan^(-1) 1 + cot^(-1) 3`
`cot^(-1) + cot^(-1).(1)/(2) + cot^(-1).(1)/(3)`
C. `cot^(-1) 1 + cot^(-1).(1)/(2) + cot^(-1).(1)/(3)`
D. `cot^(-1) 1 + tan^(-1) 3`

Explanation

To solve the series \( S = \tan^{-1}\left(\frac{4}{7}\right) + \tan^{-1}\left(\frac{4}{19}\right) + \tan^{-1}\left(\frac{4}{39}\right) + \tan^{-1}\left(\frac{4}{67}\right) + \ldots \) up to infinity, we can follow these steps:

### Step 1: Identify the general term
The general term of the series can be expressed as:
\[ \tan^{-1}\left(\frac{4}{4n^2 + 3}\right) \]
This is derived from the observation that the denominators are of the form \( 4n^2 + 3 \).

### Step 2: Rewrite the term
We can rewrite the term:
\[ \tan^{-1}\left(\frac{4}{4n^2 + 3}\right) = \tan^{-1}\left(\frac{1}{\frac{4n^2 + 3}{4}}\right) = \tan^{-1}\left(\frac{1}{n^2 + \frac{3}{4}}\right) \]

### Step 3: Use the tangent subtraction formula
Using the identity:
\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
we can express:
\[ \tan^{-1}\left(\frac{1}{n + \frac{1}{2}}\right) - \tan^{-1}\left(\frac{1}{n - \frac{1}{2}}\right) \]
This leads us to:
\[ \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right) \]

### Step 4: Sum the series
Now, we can express the sum \( S \) as:
\[ S = \sum_{n=1}^{\infty} \left( \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right) \right) \]

### Step 5: Telescoping series
This is a telescoping series. When we write out the first few terms:
\[ S = \left( \tan^{-1}\left(1 + \frac{1}{2}\right) - \tan^{-1}\left(1 - \frac{1}{2}\right) \right) + \left( \tan^{-1}\left(2 + \frac{1}{2}\right) - \tan^{-1}\left(2 - \frac{1}{2}\right) \right) + \ldots \]
Most terms will cancel, leaving us with:
\[ S = \lim_{K \to \infty} \left( \tan^{-1}\left(K + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right) \right) \]

### Step 6: Evaluate the limit
As \( K \to \infty \), \( \tan^{-1}(K + \frac{1}{2}) \to \frac{\pi}{2} \). Thus:
\[ S = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 7: Final expression
We know that:
\[ \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} - \tan^{-1}\left(2\right) \]
Thus, the final result can be expressed as:
\[ S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Conclusion
The value of the series is:
\[ S = \frac{\pi}{4} \]