I am interested in the asymptotic expansion in $t$($t>0$) when $t\to 0^+$ of the following series $$ \sum_{k\ge 0}e^{-k^{2/n}t} $$ for integer $n>2$ (n=1 follows from Poisson summation formula and n=2 is trivial). Especially the second term in the expansion.
It seems the first term in the expansion is of order $O(t^{-n/2})$, this can be obtained from estimating the sum above by the corresponding integral. But the second term seems to be much subtler, I have been trying contour integral with function $z^{n-1}e^{-z^2 t}\pi\cot(\pi z^n)$ without success.
Any references or ideas are appreciated.

Edit: I am looking at the sum since I am essentially interested in the behavior of $$ \frac{d}{dt}\left[t^{n/2}\sum_{k\ge 0}e^{-k^{2/n}t}\right] $$

when $t\to 0$. especially if it is bounded at $t=0$. I thought that if I know the second term in the expansion of $\sum_{k\ge 0}e^{-k^{2/n}t}$, then I would know the derivative.

By the comments so far I only know the coefficient in front of $t^{-n/2}$, it is good to know that(by doing contour integral I also get that), but it does not seem to tell for example if it contains a $t^{-(n-1)/2}$ as the next term in the expansion.

  • 1
    $\begingroup$ Let $\alpha=2/n$. Compare the sum in question to the integral $\int_0^\infty e^{-x^\alpha t}dt$. $\endgroup$
    – user130903
    Nov 26, 2020 at 8:46
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    $\begingroup$ Since the function $f(k)$ is bounded and decreasing, the difference between the sum $\sum f(k)$ and the integral $\int_0^\infty f(x)dx$ is bounded, so the next term is $O(1)$. What exactly do you need from it? $\endgroup$ Nov 26, 2020 at 8:58
  • $\begingroup$ @Zero This is already mentioned in OP. (Typo in your comment: the integral is against $dx$.) $\endgroup$ Nov 26, 2020 at 9:00
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    $\begingroup$ @WhiteDwarf: I think more accurate estimates are also straightforward, along these lines: en.wikipedia.org/wiki/Euler%e2%80%93Maclaurin_formula $\endgroup$ Nov 26, 2020 at 20:30
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    $\begingroup$ this derivative at t=0 equals 0, but for derivatives at $t$ close to 0 it is not enough to get an asymptotics for the sum itself $\endgroup$ Nov 26, 2020 at 23:09

1 Answer 1


The sum in question is \begin{equation} S:=\sum_0^\infty f(k):=S_1+S_2, \end{equation} where \begin{equation} f(x):=e^{-tx^a},\quad a:=2/n\in(0,1), \end{equation} \begin{equation} S_1:=\sum_0^{c-1} f(k),\quad S_2:=\sum_c^\infty f(k), \end{equation} and $c$ is an integer varying together with $t\downarrow0$ so that \begin{equation} c\to\infty,\quad tc^{1+a}\to0. \end{equation}

So, $tc^a\to0$ and hence
\begin{equation} S_1=\sum_0^{c-1}(1+O(tk^a))=c+O(tc^{1+a})=c+o(1). \end{equation}

Next, we are going to to use the Euler--Maclaurin formula. E.g., formula (2.1) in this paper or its arXiv version with $m=2$ implies \begin{equation} S_2=\int_c^\infty f(x)\,dx+\frac{f(c)}2+O(|f'(c)|)+R_2, \end{equation} where \begin{equation} |R_2|\ll\int_c^\infty |f'''(x)|\,dx; \end{equation} as usual, we write $a\ll b$ to mean $|a|=O(b)$ (the corresponding constants in $O(\cdot)$ are universal everywhere here) and $a<< b$ to mean $|a|=o(b)$. Further, \begin{align*} \int_c^\infty f(x)\,dx&=\frac1{at^{1/a}}\int_{tc^a}^\infty e^{-u}u^{1/a-1}\,du \\ &=\frac1{at^{1/a}}\,\Gamma\Big(\frac1a\Big) \\ &-\frac1{at^{1/a}}\,\int_0^{tc^a}(1+O(u))u^{1/a-1}\,du \\ &=\frac1{t^{1/a}}\,\Gamma\Big(\frac1a+1\Big)-c+O(tc^{1+a}))\\ &=\frac1{t^{1/a}}\,\Gamma\Big(\frac1a+1\Big)-c+o(1), \end{align*} \begin{equation} f(c)=e^{-tc^a}\to1, \end{equation} \begin{equation} f'(c)\ll tc^{a-1}<<tc^{a+1}<<1, \end{equation} \begin{equation} f'''(x)\ll f_3(x):=tx^{a-3}(1+t^2x^{2a})e^{-tx^a}, \end{equation} and hence \begin{equation} |R_2|\ll\int_c^\infty f_3(x)\,dx \ll t^{2/a} \int_{tc^a}^\infty e^{-u}u^{-2/a}(1+u^2)\,du \ll t^{2/a} (tc^a)^{1-2/a}=tc^{a-2}<<1. \end{equation}

Collecting all the pieces, we conclude that \begin{equation} S=\frac1{t^{1/a}}\,\Gamma\Big(\frac1a+1\Big)+\frac12+o(1). \end{equation}

Using this crucial idea with $c$ growing at an appropriate rate, and taking more terms of the Maclaurin series for the exponential function as well as more terms in the Euler--Maclaurin formula, one should be able to obtain further asymptotic expansions of the sum $S$.

  • $\begingroup$ Thanks for the answer! I think it is very helpful. $\endgroup$
    – WhiteDwarf
    Nov 27, 2020 at 21:44

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