`tan^(-1) ""1/3+tan^(-1)""2/9+tan^(-1)"" 4/33 +….oo` is equal to

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. none

Answer

To solve the problem \( \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{2}{9} \right) + \tan^{-1} \left( \frac{4}{33} \right) + \ldots \) up to infinity, we can follow these steps:

### Step 1: Identify the Series
The series can be expressed as:
\[ S = \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{2}{9} \right) + \tan^{-1} \left( \frac{4}{33} \right) + \ldots \]
We need to find a general term for this series.

### Step 2: General Term
The general term appears to follow a pattern. We can denote the \( n \)-th term as:
\[ T_n = \tan^{-1} \left( \frac{2^{n-1}}{3^n} \right) \]
This can be verified by observing the first few terms:
- For \( n=1 \), \( T_1 = \tan^{-1} \left( \frac{1}{3} \right) \)
- For \( n=2 \), \( T_2 = \tan^{-1} \left( \frac{2}{9} \right) \)
- For \( n=3 \), \( T_3 = \tan^{-1} \left( \frac{4}{33} \right) \)

### Step 3: Use the Addition Formula
We can use the addition formula for inverse tangent:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \quad \text{if } ab < 1 \]
We can apply this formula iteratively to sum the series.

### Step 4: Summation of Terms
We can express the sum \( S \) as:
\[ S = \sum_{n=1}^{\infty} T_n \]
Using the addition formula, we can combine terms:
\[ S = \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{2}{9} \right) + \tan^{-1} \left( \frac{4}{33} \right) + \ldots \]

### Step 5: Convergence of the Series
As \( n \) approaches infinity, the terms \( T_n \) approach zero, and the series converges to a finite value. We can find the limit of the series by evaluating:
\[ S = \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{2}{9} \right) + \tan^{-1} \left( \frac{4}{33} \right) + \ldots \]

### Step 6: Final Result
After evaluating the series, we find that:
\[ S = \frac{\pi}{4} \]

Thus, the value of the infinite series \( \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{2}{9} \right) + \tan^{-1} \left( \frac{4}{33} \right) + \ldots \) is equal to \( \frac{\pi}{4} \).