The sum of the infinite series `cot^(-1)(7/4)+cot^(-1)((19)/4) +cot^(-1)((39)/4)....oo`

Options

A. `(pi)/(4)-cot^(-1)(3)`
B. `(pi)/(4)-tan^(-1)(3)`
C. `(pi)/(4)+cot^(-1)(3)`
D. `(pi)/(4) +tan^(-1)(3)`

Answer

To find the sum of the infinite series \( \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \cot^{-1}\left(\frac{39}{4}\right) + \ldots \), we can follow these steps:

### Step 1: Identify the General Term
The terms in the series can be expressed as:
\[ \cot^{-1}\left(\frac{7}{4}\right), \cot^{-1}\left(\frac{19}{4}\right), \cot^{-1}\left(\frac{39}{4}\right), \ldots \]
We can observe that the numerators follow a pattern. The general term can be written as:
\[ \cot^{-1}\left(\frac{4n^2 + 3}{4}\right) \quad \text{for } n = 1, 2, 3, \ldots \]

### Step 2: Rewrite the General Term
The general term can be rewritten as:
\[ \cot^{-1}\left(\frac{4n^2 + 3}{4}\right) = \cot^{-1}(n^2 + \frac{3}{4}) \]

### Step 3: Use the Cotangent Addition Formula
Using the identity:
\[ \cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1}\left(\frac{xy - 1}{x + y}\right) \]
we can express the sum of two terms in the series.

### Step 4: Telescoping Series
We can express the series in a telescoping form:
\[ \cot^{-1}(n) - \cot^{-1}(n+1) \]
This allows us to write the series as:
\[ \sum_{n=1}^{\infty} \left(\cot^{-1}(n) - \cot^{-1}(n+1)\right) \]
which simplifies to:
\[ \cot^{-1}(1) - \lim_{n \to \infty} \cot^{-1}(n) \]
Since \( \cot^{-1}(n) \to 0 \) as \( n \to \infty \), we have:
\[ \cot^{-1}(1) - 0 = \cot^{-1}(1) = \frac{\pi}{4} \]

### Step 5: Final Result
Thus, the sum of the infinite series is:
\[ \frac{\pi}{4} \]