`overset(3)underset(n=1)Sigma tan^(-1) 1/n =`

Options

A. 0
B. `pi//2`
C. `pi`
D. none

Answer

To solve the problem \(\sum_{n=1}^{3} \tan^{-1} \frac{1}{n}\), we will evaluate the sum step by step.

### Step 1: Write the sum explicitly
We need to evaluate:
\[ \tan^{-1} \frac{1}{1} + \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} \]

### Step 2: Calculate \(\tan^{-1} \frac{1}{1}\)
\[ \tan^{-1} \frac{1}{1} = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 3: Calculate \(\tan^{-1} \frac{1}{2}\)
Let \(x = \tan^{-1} \frac{1}{2}\).

### Step 4: Calculate \(\tan^{-1} \frac{1}{3}\)
Let \(y = \tan^{-1} \frac{1}{3}\).

### Step 5: Use the formula for the sum of inverse tangents
We will use the formula:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \]
for \(xy < 1\).

First, we will find \(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}\):
\[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right)} \right) \]

### Step 6: Simplify the expression
Calculating the numerator:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \]
Calculating the denominator:
\[ 1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus,
\[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 7: Combine with \(\tan^{-1} \frac{1}{1}\)
Now we can add \(\tan^{-1} \frac{1}{1}\):
\[ \tan^{-1} \frac{1}{1} + \left(\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \]

### Final Answer
Thus, the value of the sum is:
\[ \sum_{n=1}^{3} \tan^{-1} \frac{1}{n} = \frac{\pi}{2} \]