Sharper approximation for $\sum_{k=1}^n (2k - 1)^\alpha$ for $\alpha > 0$?

Question

Consider the summation $$S_n(\alpha) := \sum_{k=1}^n (2k-1)^\alpha, \quad\mbox{where}~\alpha > 0. $$ By integration, it is possible to verify that $$ \frac{2(\alpha + 1) + (2n - 1)^{\alpha + 1} - 1}{2(\alpha + 1)}\leq S_n(\alpha)\leq \frac{(2n + 1)^{\alpha + 1} - 1}{2(\alpha + 1)} $$ However, these bounds are not very accurate. In particular, if we look at the ratio of the bounds, $$ \sup_{n \geq 1} \frac{(2n + 1)^{\alpha + 1} - 1}{2(\alpha + 1) + (2n - 1)^{\alpha + 1} - 1} \geq \frac{3^{\alpha + 1} - 1}{2(\alpha + 1) - 1} \to \infty, \quad \mbox{as}~\alpha \to \infty. $$ So the ratio of these bounds for small $n$ blows up as the exponent increases. Is it possible to have sharper bounds? In particular, it would be nice to have approximations $$ L_n(\alpha) \leq S_n(\alpha) \leq U_n(\alpha), \quad\mbox{such that}\quad \sup_{\alpha > 0} \sup_{n \geq 1} \frac{U_n(\alpha)}{L_n(\alpha)} < \infty. $$ Of course this is possible by taking $U_n = L_n = S_n$, but the goal would be to have some `simpler' $U_n, L_n$.