The sum `sum _(n=1)^ootan^(-1)(1/(2^n+2^(1-n)))` equals

Options

A. `pi/2`
B. `pi/4`
C. `pi/6`
D. `pi/3`

Answer

To solve the problem, we need to evaluate the infinite sum:

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

### Step 1: Simplify the Argument of the Arctangent
We start by simplifying the term inside the arctangent:

\[ \frac{1}{2^n + 2^{1-n}} = \frac{1}{2^n + \frac{2}{2^n}} = \frac{1}{\frac{2^{2n} + 2}{2^n}} = \frac{2^n}{2^{2n} + 2} \]

Thus, we can rewrite the sum as:

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

### Step 2: Factor the Denominator
Now, we can factor the denominator:

\[ 2^{2n} + 2 = 2(2^{2n-1} + 1) \]

So we have:

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

### Step 3: Use the Identity for the Arctangent
We can use the identity for the difference of arctangents:

\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right) \]

Let \( x = 2^n \) and \( y = 2^{n-1} \). Then:

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

This simplifies to:

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

### Step 4: Rewrite the Sum
We can express the sum \( S \) as a telescoping series:

\[ S = \frac{1}{2} \left( \tan^{-1}(2^1) - \tan^{-1}(2^0) + \tan^{-1}(2^2) - \tan^{-1}(2^1) + \tan^{-1}(2^3) - \tan^{-1}(2^2) + \ldots \right) \]

### Step 5: Evaluate the Limits
As \( n \) approaches infinity, \( \tan^{-1}(2^n) \) approaches \( \frac{\pi}{2} \) and \( \tan^{-1}(2^0) = \tan^{-1}(1) = \frac{\pi}{4} \).

Thus, we have:

\[ S = \frac{1}{2} \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8} \]

### Final Answer
The sum evaluates to:

\[ \boxed{\frac{\pi}{4}} \]