An equivalent for $\sum_{n=0}^{\infty} e^{-x\sqrt{n}}$ as $x$ tends to $0^+$ 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.
 A: 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 $\alpha>0$,

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

A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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
  $\ds{\sum_{n\ =\ 0}^{\infty}\expo{-x\root{n}} -
\bracks{{2 \over x^{2}} + \half + {\zeta\pars{3/2} \over 4\pi}\,x}}$:
  

