Find $\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$

Question

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.

Answer 1

Hint: You have $$a_k=\left(\frac{2k+1}{2k-1}\right)^{\frac1{2k+1}}\leq 1+\frac{2}{(2k-1)(2k+1)}$$ by Bernoulli's inequality.

Answer 2

We can also use AM-GM:

$$\sqrt[2k+1]{\frac{2k+1}{2k-1}}=\sqrt[2k+1]{1+\frac{2}{2k-1}} < \frac{1+\frac{2}{2k-1}+1+\ldots+1}{2k+1}$$ $$=\frac{2k+1+\frac{2}{2k-1}}{2k+1}=1+\frac{2}{(2k-1)(2k+1)}$$