I have to evaluate limit of sequence
$ \lim_{n \rightarrow\infty } \frac{n \sqrt{n} \sqrt[n]{(n+1)^{n}+n^{n+1}}}{[\sqrt n]+[2 \sqrt n]+...+[n\sqrt n]} $.
"[...]" here denotes floor function. What troubles me here the most is the floor function in it, I am not sure what to do with it.
Noting that $x-1 \lt \lfloor x \rfloor \le x$, we have $$\sum_{k=1}^{n}(k\sqrt{n}-1) \lt \sum_{k=1}^{n}\lfloor k\sqrt{n}\rfloor \le \sum_{k=1}^{n}k\sqrt{n},$$ that is, $$\frac{n(n+1)\sqrt{n}}{2}-n \lt \sum_{k=1}^{n}\lfloor k\sqrt{n}\rfloor \le \frac{n(n+1)\sqrt{n}}{2}.$$ Plug it into the limit, we get $$\frac{2(n^{n+1}+(n+1)^n)^{1/n}}{n+1} \le f(n) \lt \frac{2(n^{n+1}+(n+1)^n)^{1/n}}{n+1-2/\sqrt{n}}.$$
Since the limit of the functions on the left and right are equal, we need only to compute the one on left hand side.
\begin{align} &\lim_{n\to\infty}\frac{2(n^{n+1}+(n+1)^n)^{1/n}}{n+1} \\ =&2\lim_{n\to\infty}(1+n(\frac{n}{n+1})^n)^{1/n} \\ =&2\lim_{n\to\infty}(1+n(1-\frac{1}{n+1})^n)^{1/n} \\ =&2\lim_{n\to\infty}(1+\frac{n}{e})^{1/n} \\ =&2 \end{align}
Therefore, $\lim_{n\to\infty}f(n)=2$.