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