$\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2} = $

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

Let $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \left( \frac{n}{\sqrt{n^4+r^4}} - \frac{2 n r^2}{(n^2+r^2) \sqrt{n^4+r^4}} \right) = \frac{\pi}{k}.$ Using only the principal values of the inverse trigonometric functions,then $k^2$ is equal to:

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

Let $S = \frac{2}{1} {}^{n}C_{0} + \frac{2^{2}}{2} {}^{n}C_{1} + \frac{2^{3}}{3} {}^{n}C_{2} + \ldots + \frac{2^{n+1}}{n+1} {}^{n}C_{n}$. Then, $S$ equals

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ is equal to