Asymptotic behaviour of sum over the inverse japanese symbol

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 $c(\omega)$?

Solution 1:

If we exploit:

$$\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x^2+\omega}}-\frac{1}{x}\right)=-1+J_0(s\sqrt{\omega})=\sum_{n\geq 1}\frac{s^{2n}\omega^n(-1)^n}{4^nn!^2} \tag{1}$$ we have:

$$ c(\omega)-\gamma = \int_{0}^{+\infty}\frac{J_0(s\sqrt{\omega})-1}{e^s-1}\,ds. \tag{2} $$

Solution 2:

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 $0\le\omega\le1$: