$\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}}} = $

If $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + \dots + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {na + 1} \right) + \dots + \left( {na + n} \right)} \right]}} = \frac{1}{{60}}$ for some positive real number $a$,then $a$ is equal to

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

The value of $\lim _{n}$ ${\rightarrow \infty} \left( \frac{\sqrt{n}}{\sqrt{n^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots +\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}} \right)$ is

$\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}=$

$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$