<s>[INST] Find $\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$ 
If $n$ is a positive integer, find:
$$\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$$

I computed a few values for small $n$ and it seems it is always $n$, so I think I should prove:
$$n\le \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}} < n+1$$
The left part is easy, because each fraction is greater than $1$: 
$$\sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}} > \sum_{k=1}^n 1 = n$$
but for the right I could not prove.
 [/INST] We can also use AM-GM:
$$\sqrt[2k+1]{\frac{2k+1}{2k-1}}=\sqrt[2k+1]{1+\frac{2}{2k-1}} < \frac{1+\frac{2}{2k-1}+1+\ldots+1}{2k+1}$$
$$=\frac{2k+1+\frac{2}{2k-1}}{2k+1}=1+\frac{2}{(2k-1)(2k+1)}$$
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