Sum with binomial coefficients: $\sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m}$ with constant a

I don’t know how to give the upper bound of the following formula (the upper bound is related to $k$, and I hope the order of $k$ is a negative number)

$$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} $$ where $a\in (1/2, 1]$ is a constant. Any help in evaluating this sum would be appreciated, thanks!

I will derive an asymptotics, but without going into details. Assume that $a \geq 0$ is fixed. The main contributation to the sum will come from those $m$'s for wich $\left| {k/2 - m} \right| = o(k^{2/3} )$. In this range, we can use the normal approximation for the binomials, to deduce $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 1/2} }}{{\sqrt {k\pi } }}\sum\limits_{\left| {k/2 - m} \right| = o(k^{2/3} )} {\frac{1}{{m^{2a} }}e^{ - (k - 2m)^2 /(2k)} } $$ as $k\to +\infty$. Next, we approximate the sum by an integral and extend the range of integration to $(1,+\infty)$. This does not change the leading order behaviour. Thus, \begin{align*} \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} &\sim \frac{{2^{k + 1/2} }}{{\sqrt {k\pi } }}\int_1^{ + \infty } {\frac{1}{{x^{2a} }}e^{ - (k - 2x)^2 /(2k)} dx} \\ & = \frac{{2^{k + 1/2} }}{{\sqrt \pi }}\frac{1}{{k^{2a - 1/2} }}\int_{1/k}^{ + \infty } {\frac{1}{{t^{2a} }}e^{ - k(2t - 1)^2 /2} dt} \end{align*} as $k\to +\infty$. Now, $$ \int_{1/k}^{1/4} {\frac{1}{{t^{2a} }}e^{ - k(2t - 1)^2 /2} dt} < k^{2a} e^{ - k/8} $$ and we will see that this is negligible compared to the leading order behaviour of the full integral for large $k$. Hence, $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 1/2} }}{{\sqrt \pi }}\frac{1}{{k^{2a - 1/2} }}\int_{1/4}^{ + \infty } {\frac{1}{{t^{2a} }}e^{ - k(2t - 1)^2 /2} dt} . $$ Finally, we apply Laplace's method to the right-hand side to derive $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 2a} }}{{k^{2a} }} $$ as $k\to +\infty$.