`sum_(n=1)^(prop)tan^(-1)((4n)/(n^(4)+5))=`

Options

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

Explanation

To solve the problem \( \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) \), we can follow these steps:

### Step 1: Rewrite the Summation
We start with the summation:
\[ S = \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) \]

### Step 2: Simplify the Argument of the Arctangent
Notice that \( n^4 + 5 \) can be rewritten as \( n^4 + 1 + 4 \) to facilitate simplification:
\[ \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) = \tan^{-1}\left(\frac{4n}{n^4 + 1 + 4}\right) \]

### Step 3: Use the Identity for Arctangent
We can use the identity:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right) \]
to express the terms in the summation in a telescoping manner.

### Step 4: Consider the Form of the Terms
We can express the argument of the arctangent in a form that allows us to use the identity:
\[ \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) = \tan^{-1}\left(\frac{4n}{(n^2)^2 + 2^2}\right) \]
This suggests a relationship with the difference of two arctangents.

### Step 5: Set Up the Telescoping Series
We can express:
\[ \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) = \tan^{-1}(n^2 + 2) - \tan^{-1}(n^2 - 2) \]
This means that our summation can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \left( \tan^{-1}(n^2 + 2) - \tan^{-1}(n^2 - 2) \right) \]

### Step 6: Evaluate the Series
This series is telescoping, which means that most terms will cancel out:
\[ S = \lim_{N \to \infty} \left( \tan^{-1}(N^2 + 2) - \tan^{-1}(1^2 - 2) \right) \]
As \( N \to \infty \), \( \tan^{-1}(N^2 + 2) \to \frac{\pi}{2} \) and \( \tan^{-1}(-1) = -\frac{\pi}{4} \).

### Step 7: Final Calculation
Thus, we have:
\[ S = \frac{\pi}{2} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \]

### Conclusion
The final result of the summation is:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{4n}{n^4 + 5}\right) = \frac{3\pi}{4} \]

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