$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to

The value of $\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ is

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

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{\sqrt {{n^2} + n} }} + \frac{1}{{\sqrt {{n^2} + 2n} }} + \dots + \frac{1}{{\sqrt {{n^2} + (n - 1)n} }}} \right]$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ is equal to

$\lim _{n \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots+(2 n)^k\right]=$