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 sum

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

we can start by simplifying the argument of the inverse tangent function.

### Step 1: Simplify the Argument
The expression inside the tangent inverse can be rewritten:
\[ \tan^{-1}\left(\frac{1}{2^n + 2^{1-n}}\right) = \tan^{-1}\left(\frac{1}{2^n + \frac{2}{2^n}}\right) = \tan^{-1}\left(\frac{1}{\frac{2^n \cdot 2^n + 2}{2^n}} \right) = \tan^{-1}\left(\frac{2^n}{2^{2n} + 2}\right). \]

### Step 2: Rewrite the Sum
Thus, we can rewrite the sum as:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{2^n}{2^{2n} + 2}\right). \]

### Step 3: Use the Property of Inverse Tangent
Using the property of inverse tangent, we can express the sum as:
\[ \sum_{n=1}^{\infty} \left( \tan^{-1}(2^n) - \tan^{-1}(2^{n-1}) \right). \]

### Step 4: Telescoping Series
Notice that this is a telescoping series. When we expand it, we have:
\[ (\tan^{-1}(2^1) - \tan^{-1}(2^0)) + (\tan^{-1}(2^2) - \tan^{-1}(2^1)) + (\tan^{-1}(2^3) - \tan^{-1}(2^2)) + \ldots \]
Most terms will cancel out, and we are left with:
\[ \lim_{n \to \infty} \tan^{-1}(2^n) - \tan^{-1}(2^0). \]

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

### Step 6: Final Calculation
Thus, the sum evaluates to:
\[ \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \]

### Conclusion
The final result is:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2^n + 2^{1-n}}\right) = \frac{\pi}{4}. \]

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