I would like to obtain an equivalent form for $$ f(x)=\sum_{n=0}^{\infty} e^{-x\sqrt{n}} $$ as $x \rightarrow 0^+$. I tried without success to "remove" the $\sqrt{\cdot}$ in the summand by summing over some new index $p$ writing $\displaystyle \sum_{n=0}^{\infty} =\sum_{p=0}^{\infty}\sum_{k=p^2}^{(p+1)^2-1}$. Thanks for your help.


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 $\alpha>0$,

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


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 $S_3(1/2)$:


The correct value is $1.2849300884351988591.$

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


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.

  • $\begingroup$ Thanks for sharing these interesting results! It reminds me the indispensable book from Flajolet and Sedgewick : algo.inria.fr/flajolet/Publications/books.html $\endgroup$ – Olivier Oloa Nov 24 '14 at 20:48
  • $\begingroup$ The work of Flajolet is the defining moment of mathematical computer science in the twentieth century. Keep in mind though that the inspiration for the functional equation for $S_2(x)$ is from the paper "Mellin Transform and its Applications" by Szpankowski. $\endgroup$ – Marko Riedel Nov 24 '14 at 21:15
  • $\begingroup$ I can see that our case is considered here with Mellin transform (example 14, p. 33):algo.inria.fr/flajolet/Publications/FlGoDu95.pdf $\endgroup$ – Olivier Oloa Nov 24 '14 at 22:41
  • $\begingroup$ Thanks for the pointer which I wasn't aware of. Good to know that I got the same formula. Of course if one stopped working on problems that have been treated by Flajolet and Sedgewick we would all be out of work, their contribution is that comprehensive. $\endgroup$ – Marko Riedel Nov 24 '14 at 23:01
  • $\begingroup$ Yes, Flajolet and his team did a vast and profound work. $\endgroup$ – Olivier Oloa Nov 24 '14 at 23:48

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $\ds{\sum_{n\ =\ 0}^{\infty}\expo{-x\root{n}} - \bracks{{2 \over x^{2}} + \half + {\zeta\pars{3/2} \over 4\pi}\,x}}$: enter image description here

  • $\begingroup$ (+1). Nice work. You seem to have used the Mellin transform $$\mathfrak{M}\left(\frac{1}{e^x-1}; s\right) = \int_0^\infty \frac{x^{s-1}}{e^x-1} dx = \Gamma(s)\zeta(s).$$ We use the functional equation of the Riemann Zeta function to verify that your coefficient on $x$ is the same as mine. $\endgroup$ – Marko Riedel Nov 26 '14 at 18:29
  • $\begingroup$ Can you maybe explain how you simplified the numerator in the integral at the end of the first line? I get $$e^{-x\sqrt{it}}-e^{-x\sqrt{-it}} =e^{-x\sqrt{t}(1+i)/\sqrt{2}}-e^{-x\sqrt{t}(1-i)/\sqrt{2}} = - 2i e^{-x\sqrt{t}/\sqrt{2}} \sin(x\sqrt{t}/\sqrt{2}).$$ $\endgroup$ – Marko Riedel Nov 26 '14 at 18:52
  • $\begingroup$ @MarkoRiedel It was a typo. I just fixed it. Thanks. $\endgroup$ – Felix Marin Nov 26 '14 at 21:22
  • $\begingroup$ Do you maybe want to take another look at the integral on the second line? $\endgroup$ – Marko Riedel Nov 27 '14 at 12:30
  • $\begingroup$ @FelixMarin Nice point of view! (+1). $\endgroup$ – Olivier Oloa Nov 27 '14 at 19:33

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