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

the general term $$T_r$$ is:

$$T_r = \frac{n(n^2 - r^2)}{(n^2 + r^2)\sqrt{n^4 + r^4}}$$

Dividing the numerator and denominator by $$n^3$$:

$$T_r = \frac{1 - \frac{r^2}{n^2}}{(1 + \frac{r^2}{n^2}) \sqrt{n^2 + \frac{r^4}{n^2}}} = \frac{1 - (\frac{r}{n})^2}{(1 + (\frac{r}{n})^2) \sqrt{1 + (\frac{r}{n})^4}} \cdot \frac{1}{n}$$

2. The Trigonometric Identity

Let $$x = \frac{r}{n}$$. We can use the substitution $$x^2 = \tan \theta$$. However, to make it telescopic, we look for a form $$\tan^{-1}(A) - \tan^{-1}(B)$$.

A known identity for this specific structure involves the function:

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

This doesn't look telescopic yet, so let's try a different angle.

Consider the identity:

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

If we define a sequence $$a_r$$ such that $$T_r = a_r - a_{r-1}$$, then as $$n \to \infty$$, the sum $$\sum T_r$$ becomes $$a_n - a_0$$.

As shown in the previous step, the anti-derivative of $$g(x) = \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}$$ is:

$$G(x) = \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{\sqrt{2}x}{1+x^2} \right)$$

We can verify this by differentiating $$G(x)$$:

$$\frac{d}{dx} \left[ \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{\sqrt{2}x}{1+x^2} \right) \right] = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{1 - \left(\frac{\sqrt{2}x}{1+x^2}\right)^2}} \cdot \frac{d}{dx} \left( \frac{\sqrt{2}x}{1+x^2} \right)$$

$$= \frac{1}{\sqrt{2}} \cdot \frac{1+x^2}{\sqrt{(1+x^2)^2 - 2x^2}} \cdot \sqrt{2} \left( \frac{1-x^2}{(1+x^2)^2} \right)$$

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

The sum telescopes across the interval $$[0, 1]$$:

$$L = G(1) - G(0)$$

$$L = \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{\sqrt{2}(1)}{1+1^2} \right) - \frac{1}{\sqrt{2}} \sin^{-1}(0)$$

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

This gives us:

• $$k = 4\sqrt{2}$$
• $$k^2 = (4\sqrt{2})^2 = \mathbf{32}$$