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

<p>$$\begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\ & -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}} \\ & =\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n \sqrt{1+\frac{r^4}{n^4}}}-\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n} \frac{2 \cdot(r / n)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\frac{r^4}{n^4}}} \\ & =\int_\limits0^1 \frac{d x}{\sqrt{1+x^4}}-2 \int_\limits0^1 \frac{x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\ & =\int_\limits0^1 \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \end{aligned}$$</p> <p>$$\begin{aligned} & \int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} \\ & =\int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} \\ & x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t \\ & -\int_\limits{\infty}^2 \frac{d t}{t \sqrt{t^2-2}}=\int_\limits2^{\infty} \frac{d t}{t \sqrt{t^2-2}}=\left.\frac{-1}{\sqrt{2}} \sin ^{-1} \frac{\sqrt{2}}{x}\right|_2 ^{\infty} \\ & =\frac{-1}{\sqrt{2}}\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{2^{5 / 2}} \\ & \therefore k=2^{5 / 2} \\ & \therefore k^2=32 \end{aligned}$$</p>