$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {3n} \right)}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} = $

Evaluate $\int_{0}^{2} e^{x} dx$ as the limit of a sum.

$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$

$\mathop {Lim}\limits_{n \to \infty } \frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2 \cdot \frac{\pi }{{6n}}} \right) + \dots + {{\sec }^2}\left( {(n - 1)\frac{\pi }{{6n}}} \right) + \frac{4}{3}} \right]$ has the value equal to

$\lim _{n \rightarrow \infty}\left[\frac{1^k+2^k+3^k+\ldots+n^k}{n^{k+1}}\right]=$

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