$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $ is equal to

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}}\left[ {1\cos \frac{1}{{{n^2}}} + 2\cos \frac{4}{{{n^2}}} + 3\cos \frac{9}{{{n^2}}} + .... + 2n\cos 4} \right]$ is equal to-

$\lim _{n \rightarrow \infty} \frac{1}{2^{n}}\left(\frac{1}{\sqrt{1-\frac{1}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{2}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{3}{2^{n}}}}+\ldots+\frac{1}{\sqrt{1-\frac{2^{n}-1}{2^{n}}}}\right)$ is equal to

$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots \left(1+\frac{n^3}{n^3}\right)^{\frac{n^2}{n^3}}\right]=$

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\left[ {{1^2}\sin \frac{1}{n} + {2^2}\sin \frac{2}{n} + {3^2}\sin \frac{3}{n} + ....+{n^2}\sin \frac{n}{n}} \right]$ equals

The value of $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{n}{{1 + {n^2}}} + \frac{n}{{4 + {n^2}}} + \frac{n}{{9 + {n^2}}} + .... + \frac{1}{{2n}}} \right]$ is equal to