An equivalent for $\sum_{n=0}^{\infty} e^{-x\sqrt{n}}$ as $x$ tends to $0^+$

It seems to have escaped attention that this sum may be evaluated using harmonic summation techniques which can be an instructive exercise.

Introduce $S_\alpha(x)%$ given by $$S_\alpha(x) = \sum_{n\ge 1} \exp(-(nx)^\alpha).$$

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \exp(-x^\alpha).$$ We need the Mellin transform $g^*(s)$ of $g(x)$, which is

$$\int_0^\infty e^{-x^\alpha} x^{s-1} dx = \int_0^\infty e^{-t} t^{(s-1)/\alpha} \frac{1}{\alpha} t^{1/\alpha-1} dt \\ = \frac{1}{\alpha} \int_0^\infty e^{-t} t^{s/\alpha-1} dt = \frac{1}{\alpha} \Gamma(s/\alpha).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \frac{1}{\alpha} \Gamma(s/\alpha) \zeta(s) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^s} = \zeta(s)$$ for $\Re(s) > 1.$

The fundamental strip of the transform of the base function is $\langle 0, \infty \rangle$ and intersecting this with $\langle 1, \infty\rangle$ we obtain that the Mellin inversion integral here is

$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

There are two types of poles: the one from the zeta function at $s=1$ and the ones from the gamma function at $-q\alpha,$ where $q\ge 0.$

We have $$\mathrm{Res}\left(Q(s)/x^s; s=1\right) = \frac{1}{\alpha}\Gamma(1/\alpha) \times \frac{1}{x} = \frac{\Gamma(1+1/\alpha)}{x}$$ and $$\mathrm{Res}\left(Q(s)/x^s; s=-q\alpha\right) = \frac{1}{\alpha} \times \alpha \frac{(-1)^q}{q!} \times \zeta(-q\alpha) \times x^{q\alpha} \\ = \frac{(-1)^q}{q!} \zeta(-q\alpha) \times x^{q\alpha}.$$

Now there are two cases, either $\alpha$ is an even integer or not. If it is, all the poles for $q>0$ are canceled by the zeta function term, leaving just the poles at $s=1$ and at $s=0$ for a result of $$ S_\alpha(x) \sim \frac{\Gamma(1+1/\alpha)}{x} - \frac{1}{2}.$$

When $\alpha = 2$ we get $$S_2(x) = - \frac{1}{2} + \frac{\sqrt{\pi}}{2x} + \frac{1}{2\pi i} \int_{-1-i\infty}^{-1+i\infty} Q(s)/x^s ds.$$

Substitute $s=1-t$ in the remainder integral to get $$\frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} Q(1-t)/x^{1-t} dt = \frac{1}{x} \frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} \frac{1}{2} \Gamma\left(\frac{1-t}{2}\right) \zeta(1-t) x^t dt.$$

Recall the following variant of the functional equation of the Riemann Zeta function: $$\Gamma\left(\frac{1-t}{2}\right) \zeta(1-t) = \pi^{1/2-t} \Gamma\left(\frac{t}{2}\right) \zeta(t)$$ and substitute it into the remainder integral to get $$\frac{1}{x} \frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} \frac{1}{2} \pi^{1/2-t} \Gamma\left(\frac{t}{2}\right) \zeta(t) x^t dt = \frac{\sqrt{\pi}}{x} \frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} \frac{1}{2} \Gamma\left(\frac{t}{2}\right) \zeta(t) \left(\frac{x}{\pi}\right)^t dt.$$

Shift this to $\Re(t)=3/2$ (no poles to pick up) and obtain $$\frac{\sqrt{\pi}}{x} \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{1}{2} \Gamma\left(\frac{t}{2}\right) \zeta(t) \left(\frac{x}{\pi}\right)^t dt = \frac{\sqrt{\pi}}{x} S_2(\pi/x).$$

This yields the functional equation $$S_2(x) = - \frac{1}{2} + \frac{\sqrt{\pi}}{2x} + \frac{\sqrt{\pi}}{x} S_2(\pi/x).$$ which is a variant of Jacobi's theta function identity.

Returning to the main thread we observe that when $\alpha$ is not an even integer we have the expansion

$$ S_\alpha(x) \sim \frac{\Gamma(1+1/\alpha)}{x} - \frac{1}{2} + \sum_{q\ge 1} \frac{(-1)^q}{q!} \zeta(-q\alpha) \times x^{q\alpha}.$$

The study of the convergence properties of this expansion is a delicate matter and certainly a wonderful challenge. E.g. with $\alpha$ an odd integer we get a non-zero contribution from the zeta function terms but we have $$\zeta(-q\alpha) = - \frac{B_{q\alpha+1}}{q\alpha + 1} \quad\text{and}\quad |B_{2n}| \sim 4\sqrt{\pi n} \left(\frac{n}{\pi e}\right)^{2n}$$

so the zeta function values outgrow the factorial in $q$ and the expansion eventually diverges.

The following table illustrates the process when calculating $S_3(1/2)$:

1.285959023138498e+00
1.284917356471832e+00
1.284919822538751e+00
1.284919709811953e+00
1.284919736441052e+00
1.284919716423841e+00
1.284919753275291e+00
1.284919612356084e+00
1.284920611708108e+00
1.284908513429109e+00
1.285143226270321e+00
1.278200633277179e+00
1.578934319042731e+00
-1.687446462625181e+01
1.542969266697830e+03
-1.758244058759777e+05
2.640441955632738e+07
-5.130529386490404e+09
1.270035750847759e+12
-3.950618113631400e+14
1.525356085967402e+17

The correct value is $1.2849300884351988591.$

Similarly when computing $S_5(1/3)$ we get the table

2.254506227199282e+00
2.254522557462526e+00
2.254522552313936e+00
2.254522552853181e+00
2.254522551342709e+00
2.254522594231955e+00
2.254515285982698e+00
2.259889649827273e+00
-1.133507270756404e+01
1.002166612666027e+05
-1.897690119553089e+09
8.353404075737941e+13
-7.886089945242315e+18
1.493955388567194e+24
-5.370850993208232e+29
3.494071363049499e+35
-3.948314085212110e+41
7.478102802029752e+47
-2.300704020425045e+54
1.118323430713735e+61
-8.378457528967820e+67

The correct value is $2.2551776136424966417$.

To conclude we recall that the sum proposed for evaluation was slightly different, namely $$f_\alpha(x) = \sum_{n\ge 0} \exp(-x n^\alpha).$$ This is readily seen to be equal to $$S_\alpha(x^{1/\alpha}) + 1$$ and hence $$f_\alpha(x) \sim \frac{\Gamma(1+1/\alpha)}{x^{1/\alpha}} + \frac{1}{2} + \sum_{q\ge 1} \frac{(-1)^q}{q!} \zeta(-q\alpha) \times x^q.$$

There is a similar calculation at this MSE link.

As $x$ tends to $0^+$, the series goes to infinity and we have

$$f(x)=\sum_{n=0}^{+\infty}e^{-x\sqrt{n}} \sim \frac{2}{x^2}. \tag{*}$$

Proof. Let $n\geq 1$. Since $\displaystyle (-\infty,0]\ni t \rightarrow e^{-x\sqrt{t}}$ is a decreasing function, we have $$ e^{-x\sqrt{n+1}} \leq e^{-x\sqrt{t}} \leq e^{-x\sqrt{n}}, \quad t \in [n,n+1], \tag1 $$ integrating $(1)$, we get $$ \int_n^{n+1}e^{-x\sqrt{t}}dt \leq e^{-x\sqrt{n}} \tag2 $$ and $$ e^{-x\sqrt{n}} \leq \int_{n-1}^{n}e^{-x\sqrt{t}} dt. \tag3 $$ Then, summing $(2)$ for $n\geq0$, gives $$ \int_0^{+\infty}e^{-x\sqrt{t}}dt \leq \sum_{n=0}^{+\infty}e^{-x\sqrt{n}} \tag4 $$ and summing $(3)$ for $n\geq 1$, gives $$ \sum_{n=0}^{+\infty}e^{-x\sqrt{n}} \leq 1+\int_0^{+\infty}e^{-x\sqrt{t}}dt. \tag5 $$By the change of variable $u=\sqrt{t},$ $t=u^2,$ $dt=2udu$, we readily have $$ \int_0^{+\infty}e^{-x\sqrt{t}}dt=2\int_0^{+\infty}ue^{-xu}du = \frac{2}{x^2} .\tag6 $$ Hence combining $(4)$, $(5)$ and $(6)$, leads to the desired result $(*)$.

Remark. The same reasoning shows that, for $\alpha>0$,

$$ f_{\alpha}(x)=\sum_{n=0}^{+\infty}e^{\large-xn^{\alpha}} \sim_{0^+} \frac{\Gamma(1+1/\alpha)}{x^{1/\alpha}}. $$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With Abel-Plana Formula :

\begin{align}&\color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\expo{-x\root{n}}} \\[5mm]&=\ \overbrace{\int_{0}^{\infty}\expo{-x\root{t}}\,\dd t} ^{\ds{\color{#c00000}{\root{t}\ \mapsto\ t}}}\ +\ \left.\half\,\expo{-x\root{t}}\right\vert_{\, t\ =\ 0} +\ic\int_{0}^{\infty} {\expo{-x\root{\ic t}} - \expo{-x\root{-\ic t}} \over \expo{2\pi t} - 1}\,\dd t \\[5mm]&=2\ \overbrace{\int_{0}^{\infty}\expo{-xt}t\,\dd t} ^{\ds{\color{#c00000}{1 \over x^{2}}}}\ +\ \half - 2\Im\int_{0}^{\infty} {\exp\pars{-x\root{t}\pars{1 + \ic}/\root{2}} \over \expo{2\pi t} - 1}\,\dd t \\[5mm]&={2 \over x^{2}} + {1 \over 2} +2\int_{0}^{\infty}\exp\pars{-\,{x \over \root{2}}\,\root{t}}\,{\sin\pars{\root{2}x\root{t}/2} \over \expo{2\pi t} - 1}\,\dd t \\[5mm]&\sim\color{#66f}{\large{2 \over x^{2}} + {1 \over 2} +\ \underbrace{\pars{\root{2}\int_{0}^{\infty}{\root{t} \over \expo{2\pi t} - 1}\,\dd t}}_{\ds{\color{#c00000}{{\zeta\pars{3/2} \over 4\pi}\ \color{#000}{\approx\ 0.2079}}}}\ x \quad\mbox{when}\quad x \sim 0} \end{align}

$$ \color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\expo{-x\root{n}} \sim {2 \over x^{2}}\,, \qquad x \sim 0} $$

Here, we can see a plot of the difference $\ds{\sum_{n\ =\ 0}^{\infty}\expo{-x\root{n}} - \bracks{{2 \over x^{2}} + \half + {\zeta\pars{3/2} \over 4\pi}\,x}}$: