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.
 A: 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 $S$.
