# An asymptotic expansion of a infinite sum

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.

• Let $\alpha=2/n$. Compare the sum in question to the integral $\int_0^\infty e^{-x^\alpha t}dt$. – Zero Nov 26 '20 at 8:46
• 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? – Fedor Petrov Nov 26 '20 at 8:58
• @Zero This is already mentioned in OP. (Typo in your comment: the integral is against $dx$.) – Fedor Petrov Nov 26 '20 at 9:00
• @WhiteDwarf: I think more accurate estimates are also straightforward, along these lines: en.wikipedia.org/wiki/Euler%e2%80%93Maclaurin_formula – Christian Remling Nov 26 '20 at 20:30
• 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 – Fedor Petrov Nov 26 '20 at 23:09

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

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

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 $$$$S_2=\int_c^\infty f(x)\,dx+\frac{f(c)}2+O(|f'(c)|)+R_2,$$$$ where $$$$|R_2|\ll\int_c^\infty |f'''(x)|\,dx;$$$$ as usual, we write $$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)$$. Further, \begin{align*} \int_c^\infty f(x)\,dx&=\frac1{at^{1/a}}\int_{tc^a}^\infty e^{-u}u^{1/a-1}\,du \\ &=\frac1{at^{1/a}}\,\Gamma\Big(\frac1a\Big) \\ &-\frac1{at^{1/a}}\,\int_0^{tc^a}(1+O(u))u^{1/a-1}\,du \\ &=\frac1{t^{1/a}}\,\Gamma\Big(\frac1a+1\Big)-c+O(tc^{1+a}))\\ &=\frac1{t^{1/a}}\,\Gamma\Big(\frac1a+1\Big)-c+o(1), \end{align*} $$$$f(c)=e^{-tc^a}\to1,$$$$ $$$$f'(c)\ll tc^{a-1}< $$$$f'''(x)\ll f_3(x):=tx^{a-3}(1+t^2x^{2a})e^{-tx^a},$$$$ and hence $$$$|R_2|\ll\int_c^\infty f_3(x)\,dx \ll t^{2/a} \int_{tc^a}^\infty e^{-u}u^{-2/a}(1+u^2)\,du \ll t^{2/a} (tc^a)^{1-2/a}=tc^{a-2}<<1.$$$$

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

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 $$S$$.

• Thanks for the answer! I think it is very helpful. – WhiteDwarf Nov 27 '20 at 21:44