TITLE: Asymptotic behaviour of sum over the inverse japanese symbol
QUESTION [10 upvotes]: I am interested in the asymptotic behavior of the sum
$$\sum_{m=1}^M\frac{1}{\sqrt{m^2+\omega}}$$
for $1>\omega>0$ in the Limit $M\to\infty$ up to order $\mathcal{O}(M^{-1})$.
The first thing I did was splitting the sum as follows:
$$\sum_{m=1}^M\frac{1}{\sqrt{m^2+\omega}}=\sum_{m=1}^M\frac{1}{m}+\sum_{m=1}^M\left(\frac{1}{\sqrt{m^2+\omega}}-\frac{1}{m}\right)$$
For the first sum I use the Euler approximation to get
$$\sum_{m=1}^M\frac{1}{m}=\log(M)+\gamma_E+\mathcal{O}(M^{-1}).$$
Now I suspect that the remainder goes like
$$\sum_{m=1}^M\left(\frac{1}{\sqrt{m^2+\omega}}-\frac{1}{m}\right)=c(\omega)+\mathcal{O}(M^{-1}).$$
The question is, how can I explicitly compute $c(\omega)$?

REPLY [4 votes]: The Euler-Maclaurin Sum Formula applied to the power series in $\frac1m$ yields

$$
\begin{align}
\sum_{m=1}^M\frac1{\sqrt{m^2+\omega}}
&=\log(M)+c(\omega)+\frac1{2M}+\frac{3\omega-1}{12M^2}-\frac\omega{4M^3}\\
&+\frac{4+60\omega-45\omega^2}{480M^4}+\frac{3\omega^2}{M^5}+O\!\left(\frac1{M^6}\right)\tag{1}
\end{align}
$$

and
$$
\sum_{m=1}^M\frac1m
=\log(M)+\gamma+\frac1{2M}-\frac1{12M^2}+\frac1{120M^4}+O\!\left(\frac1{M^6}\right)\tag{2}
$$
Note that as $\omega\to0$, $(1)\to(2)$, as long as $c(\omega)\to\gamma$.
Subtracting $(2)$ from $(1)$ and letting $M\to\infty$, we get
$$
c(\omega)-\gamma
=\sum_{m=1}^\infty\left(\frac1{\sqrt{m^2+\omega}}-\frac1m\right)\tag{3}
$$
Therefore,
$$
\begin{align}
c(\omega)
&=\gamma+\sum_{m=1}^\infty\left(\frac1{\sqrt{m^2+\omega}}-\frac1m\right)\\
&=\gamma+\frac1{\sqrt{1+\omega}}-1+\sum_{m=2}^\infty\left(\frac1{\sqrt{m^2+\omega}}-\frac1m\right)\\
&=\gamma+\frac1{\sqrt{1+\omega}}-1+\sum_{m=2}^\infty\frac1m\sum_{k=1}^\infty\binom{2k}{k}\left(-\frac\omega{4m^2}\right)^k\\
&=\gamma+\frac1{\sqrt{1+\omega}}-1+\sum_{k=1}^\infty\left(-\frac\omega4\right)^k\binom{2k}{k}\underbrace{(\zeta(2k+1)-1)}_{\sim2^{-2k-1}}\tag{4}
\end{align}
$$
That is,

$$
c(\omega)=\gamma+\frac1{\sqrt{1+\omega}}-1+\sum_{k=1}^\infty\left(-\frac\omega4\right)^k\binom{2k}{k}(\zeta(2k+1)-1)\tag{5}
$$

The absolute value of the coefficient of $\omega^k$ in the summation is $\,\sim\!\frac1{4^k\sqrt{4\pi k}}$, so the sum converges for $\left|\omega\right|\lt4$. Therefore, $\lim\limits_{\omega\to0}c(\omega)=\gamma$, as required above.

Note that if we incorporate $\frac1{\sqrt{1+\omega}}-1$ into the summation in $(4)$ and move $\gamma$ to the right side, we get
$$
c(\omega)-\gamma=\sum_{k=1}^\infty\left(-\frac\omega4\right)^k\binom{2k}{k}\zeta(2k+1)\tag{6}
$$
which matches formula $(1)$ from Jack D'Aurizio's answer applied to the integral in $(2)$ from Jack D'Aurizio's answer. However, $(6)$ only converges for $-1\lt\omega\le1$.

Here is a plot of $c(\omega)$ for $0\le\omega\le1$: