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

We start with the given series:

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

Simplify the expression:

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

Let’s use the substitution \( x = \tan \theta \):

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

Substituting these values, we get:

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

\[ = \sin \theta + C \]

Now reverting back to \( x \):

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

Next, apply the limits:

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

Thus,

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

Answer 2

The given sum is:

$$\sum_{r=1}^{\infty} \left(\frac{n}{\sqrt{n^4 + r^4}} - \frac{2nr^2}{(n^2 + r^2)\sqrt{n^4 + r^4}}\right)$$

Substitute \(r = \frac{r}{n}\) and simplify:

$$\sum_{r=1}^{\infty} \left(\frac{1}{n} \cdot \frac{1}{\sqrt{1 + \left(\frac{r}{n}\right)^4}} - \frac{2\left(\frac{r}{n}\right)^2}{\left(1 + \left(\frac{r}{n}\right)^2\right)\sqrt{1 + \left(\frac{r}{n}\right)^4}}\right)$$

In the limit as \(n \to \infty\), this becomes the integral:

$$\int_0^1 \frac{dx}{\sqrt{1 + x^4}} - \int_0^1 \frac{2x^2dx}{(1 + x^2)\sqrt{1 + x^4}}$$

Simplify:

$$\int_0^1 \frac{1 - x^2}{(1 + x^2)\sqrt{1 + x^4}} dx$$

Substitute \(x + \frac{1}{x} = t\), where \(1 - \frac{1}{x^2} dx = dt\). The integral becomes:

$$\int_{\sqrt{2}}^{\infty} \frac{dt}{t\sqrt{t^2 - 2}}$$

Now substitute \(t^2 - 2 = \alpha^2\), so \(tdt = \alpha d\alpha\). The integral becomes:

$$\int_0^{\infty} \frac{\alpha d\alpha}{(\alpha^2 + 2)\alpha} = \int_{\sqrt{2}}^{\infty} \frac{d\alpha}{\alpha^2 + 2}$$

This simplifies further using the inverse tangent:

$$\int_{\sqrt{2}}^{\infty} \frac{d\alpha}{\alpha^2 + 2} = \frac{1}{\sqrt{2}} \left[\tan^{-1} \frac{\alpha}{\sqrt{2}}\right]_{\sqrt{2}}^{\infty}$$

At the limits:

$$\frac{1}{\sqrt{2}} \left(\tan^{-1} \infty - \tan^{-1} 1\right) = \frac{1}{\sqrt{2}} \left(\frac{\pi}{2} - \frac{\pi}{4}\right)$$

Simplify:

$$= \frac{\pi}{4\sqrt{2}}$$

We are given \(\frac{\pi}{k}\), so:

$$K = 4\sqrt{2}$$

Thus:

$$K^2 = 32$$

Final Answer is 32