$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ is equal to

By the definition of the definite integral, the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

$\mathop {Limit}\limits_{n \to \infty } \frac{1}{n} \left[ 1 + \sqrt {\frac{n}{n + 1}} + \sqrt {\frac{n}{n + 2}} + \sqrt {\frac{n}{n + 3}} + \dots + \sqrt {\frac{n}{n + 3(n - 1)}} \right]$ has the value equal to

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

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals