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

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

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