Sum the series : `tan^(- 1)(4/(1+3.4))+tan^(- 1)(6/(1+8.9))+tan^(- 1)(8/(1+15.16))+............oo` is :

Options

A. `cot^(-1)(2)`
B. `tan^(-1)(2)`
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To find the sum of the series given by

\[ S = \tan^{-1}\left(\frac{4}{1 + 3 \cdot 4}\right) + \tan^{-1}\left(\frac{6}{1 + 8 \cdot 9}\right) + \tan^{-1}\left(\frac{8}{1 + 15 \cdot 16}\right) + \ldots \]

we will first identify the general term of the series.

### Step 1: Identify the General Term

The general term can be expressed as:

\[ T_r = \tan^{-1}\left(\frac{2r + 2}{1 + (r + 1)(r + 2)}\right) \]

where \( r \) starts from 1 and goes to infinity.

### Step 2: Simplify the Denominator

The denominator can be simplified as follows:

\[ 1 + (r + 1)(r + 2) = 1 + (r^2 + 3r + 2) = r^2 + 3r + 3 \]

Thus, we can rewrite the general term as:

\[ T_r = \tan^{-1}\left(\frac{2r + 2}{r^2 + 3r + 3}\right) \]

### Step 3: Rewrite the General Term

Now, we can express \( T_r \) in a more manageable form:

\[ T_r = \tan^{-1}\left(\frac{2(r + 1)}{(r + 1)(r + 3)}\right) = \tan^{-1}\left(\frac{2}{r + 3}\right) \]

### Step 4: Sum the Series

The series can now be rewritten as:

\[ S = \sum_{r=1}^{\infty} \left( \tan^{-1}\left(\frac{2}{r + 3}\right) \right) \]

### Step 5: Use the Inverse Tangent Identity

Using the identity for the difference of inverse tangents, we can express the series as:

\[ S = \tan^{-1}(2) - \tan^{-1}(0) \]

### Step 6: Evaluate the Limit

As \( r \to \infty \), \( \tan^{-1}(0) \) approaches 0. Thus, the sum converges to:

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

### Step 7: Final Result

The final result for the sum of the series is:

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

### Conclusion

Thus, the sum of the series is:

\[ \text{Sum} = \cot^{-1}(2) \]