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!
 A: 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 $k\to +\infty$.
