[Timestopper_STG]

I also solved it, but just for all real ${\alpha}\geq 0$...

At first, when $\alpha=0$, then we have $S_{\alpha}(n)=1,\forall n\in\mathbb{N}$

Consider $\alpha >0$.

Notice that in this case $\displaystyle{A_{\alpha}(n)<B_{\alpha}(n),\forall n\in\mathbb{N}}$.

Hence, it follows that

$$1-\frac{(2n)^{\alpha}}{B_{\alpha}(n)}=\frac{0^{\alpha}+2^{\alpha}+\cdots+(2n-2)^{\alpha}}{B_{\alpha}(n)}<S_{\alpha}(n)=\frac{A_{\alpha}(n)}{B_{\alpha}(n)}<\frac{2^{\alpha}+4^{\alpha}+\cdots+(2n)^{\alpha}}{ B_{\alpha}(n)}=1$$

Then observe that

$$\lim_{n \rightarrow \infty} \frac{B_{\alpha}(n)}{(2n)^{\alpha}}= \lim_{ n \rightarrow \infty}\frac{1^{\alpha}+2^{\alpha}+\cdots + n^{\alpha}}{n^{\alpha}}=\lim_{n \rightarrow \infty}n\left[ \frac{1}{n}\sum_{k=1}^{n}f\left( \frac{k}{n} \right) \right]$$

So look at the limit of the first inequality we have $\displaystyle{\lim_{n\rightarrow\infty}S_{\alpha}(n)=1}$.