Question

Let \(\lim_{n \to \infty} \left( \frac{n}{\sqrt{n^4 + 1}} - \frac{2n}{\left(n^2 + 1\right)\sqrt{n^4 + 1}} + \frac{n}{\sqrt{n^4 + 16}} - \frac{8n}{\left(n^2 + 4\right)\sqrt{n^4 + 16}} + \ldots + \frac{n}{\sqrt{n^4 + n^4}} - \frac{2n \cdot n^2}{\left(n^2 + n^2\right)\sqrt{n^4 + n^4}} \right)\)  be \(\frac{\pi}{k},\) using only the principal values of the inverse trigonometric functions. Then \(k^2\) is equal to ______.

Answer 1

Correct Answer: 32

The provided series is:

\[ \sum \frac{1}{\sqrt{x^2 + r^2}} = \int \frac{2x^2}{(x^2 + r^2)\sqrt{1 + (x^2 + r^2)}} \]

The expression is simplified as:

\[ \int \frac{dx}{\sqrt{1 + x^2}(1 + x^2)} = \int \frac{2x \, dx}{(1 + x^2)\sqrt{1 + x^2}} \]

Using the substitution \( x = \tan \theta \):

\[ dx = \sec^2 \theta \, d\theta, \quad \sqrt{1 + x^2} = \sec \theta \]

After substituting these values:

\[ \int \frac{\sec^2 \theta \, d\theta}{\sec^3 \theta} = \int \cos \theta \, d\theta \]

\[ = \sin \theta + C \]

Returning to the variable \( x \):

\[ \sin \theta = \frac{x}{\sqrt{1 + x^2}} \Rightarrow \int \frac{dx}{(x^2 + 1)\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}} + C \]

Applying the limits:

\[ \left[\tan^{-1}\left(\frac{x}{r}\right)\right]_0^\infty = \frac{\pi}{2r} \]

Consequently:

\[ K = 4\sqrt{2}, \quad \text{and} \quad K^2 = 32 \]