`sum_(n=1)^oo(tan^-1((4n)/(n^4-2n^2+2)))` is equal to (A) `tan ^-1 (2)+tan^-1 (3)` (B) `4tan^-1 (1)` (C) `pi/2` (D) `sec^-1(-sqrt2)`

Options

A. `tan^(-1). 1/2 + tan^(-1). 2/3`
B. `4 tan^(-1) 1`
C. `pi/2`
D. `sec^(-1) ( - sqrt2)`

Answer

To solve the given infinite series \( \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{4n}{n^4 - 2n^2 + 2}\right) \), we will follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the summation:
\[ \tan^{-1}\left(\frac{4n}{n^4 - 2n^2 + 2}\right) \]
We can rewrite the denominator:
\[ n^4 - 2n^2 + 2 = (n^2 - 1)^2 + 1 \]
Thus, we have:
\[ \tan^{-1}\left(\frac{4n}{(n^2 - 1)^2 + 1}\right) \]

### Step 2: Use the Identity for Inverse Tangent
We can express \( \frac{4n}{(n^2 - 1)^2 + 1} \) in a form suitable for the tangent subtraction formula:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
We can rewrite \( 4n \) as:
\[ 4n = (n+1)^2 - (n-1)^2 \]
This allows us to express the argument as:
\[ \tan^{-1}\left(\frac{(n+1)^2 - (n-1)^2}{(n^2 - 1)^2 + 1}\right) \]

### Step 3: Apply the Inverse Tangent Formula
Using the identity, we can express:
\[ \tan^{-1}(n+1) - \tan^{-1}(n-1) \]
Thus, we can rewrite our series:
\[ \sum_{n=1}^{\infty} \left( \tan^{-1}(n+1) - \tan^{-1}(n-1) \right) \]

### Step 4: Recognize the Telescoping Series
Notice that this is a telescoping series:
\[ \sum_{n=1}^{\infty} \left( \tan^{-1}(n+1) - \tan^{-1}(n-1) \right) \]
This series will cancel out many terms:
- The first term from \( n=1 \) gives \( \tan^{-1}(2) - \tan^{-1}(0) \)
- The last term as \( n \to \infty \) gives \( \tan^{-1}(\infty) - \tan^{-1}(0) \)

### Step 5: Evaluate the Limits
As \( n \to \infty \), \( \tan^{-1}(n+1) \to \frac{\pi}{2} \) and \( \tan^{-1}(0) = 0 \). Thus, we have:
\[ \lim_{n \to \infty} \left( \tan^{-1}(n+1) - \tan^{-1}(n-1) \right) = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Conclusion
The value of the infinite series is:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{4n}{n^4 - 2n^2 + 2}\right) = \frac{\pi}{2} \]

### Final Answer
Thus, the answer is \( \frac{\pi}{2} \), which corresponds to option (C).