I am trying to prove or disprove

$$\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} ,$$

where $\sum c_{k}<\infty, \sum c_{k}^{2}<\infty\text{ and }\frac{\lambda_{k}}{k}\to c$(Weyl's law). The $c_{k},\lambda_{k},t\in \mathbb{R}$ and

**Update**: the $\lambda_{k}$ are eigenvalues of any domain D in $\mathbb{R}^{2}$:

- $\lambda_{k}<\lambda_{k+1}<....$
- $\lambda_{k}=\frac{4\pi}{|D|}k+c_{2}\sqrt{k}+o(\sqrt{k})$
- $\sum^{k}_{j\geq 1}\lambda_{j}\geq \frac{2\pi}{|D|}k^{2}\Rightarrow \lambda_{k}\geq \frac{2\pi}{|D|}k$ (Li-Yau)

The counterexamples below were before the update, where we only assumed $\frac{\lambda_{k}}{k}\to c$ (but they seem to include monotonicity too).

Q:Since the line of convergence is $Re(z)=0$, is there any way to analytic continue this series in a neighbourhood of zero, in order to apply some of the known tauberian theorems.

Please, I prefer just hints to help me practice.

Attempts

0)The $\lambda_{k}$ correspond to the Dirichlet-Laplacian eigenvalues of a domain in $\mathbb{R}^{2}$, so I was hoping to have them fixed and only modify the $c_{k}$ for a counterexample. Or come up with the $\lambda_{k}$ and the corresponding domain, that gives a counterexample.

1)The proof of Abel's theorem doesn't work because it requires linear error term $|\lambda_{k}/k-c|<\frac{c'}{k}$ (from Weyl's law we have $\frac{1}{\sqrt{k}}$ error): for $|s_{N}|=|\sum_{N}c_{k}|\leq \varepsilon$ we have $|\sum_{N} c_{k} e^{-\lambda_{k}t}|\leq \varepsilon \sum_{N} ( e^{-\lambda_{k}t}- e^{-\lambda_{k+1}t})\approx \varepsilon e^{-N c_{1} t}\frac{(1-e^{-\sqrt{N+1}t c_{2}})}{(1-e^{t c_{2}})}\to \varepsilon c \sqrt{N+1}$ assuming the unproved but plausible $\lambda_{k}\geq \frac{4\pi}{|D|}k$ for large k.

2)Littlewood tauberian theorems requiring $c_{k}k=o(1)$ don't apply because we might have $c_{k}=\frac{(-1)^{k}}{k}$.

3)By having $c_{k}=\frac{(-1)^{k}}{k}$ the line of convergence of abstract Dirichlet series

$$\sum e^{-\lambda z}c_{k}$$. So analytic continuation is not clear.

is $\{z:Re(z)=0\}$. The Ostrowski–Hadamard gap theorem doesn't apply because $\lambda_{k+1}/\lambda_{k}\to 1$.

4)Borel summation method might apply: Because $\sum c_{k}^{2}<\infty$ and so $$\sqrt{k}c_{k}=o(1).$$

Moreover, $e^{-x}\sum s_{k}\frac{1}{k!}z^{k}$ is weakly Borel summable. So indeed

$$\sum_{k=1}^{\infty}e^{-kt}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k}.$$

So maybe the proof can be modified for $\lambda_{k}$ instead.

5)Possible counterexample via letting $c_{k}=\frac{(-1)^{k}}{k}$ to disallow any dominated convergence.

6)Another Tauberian theorem requires showing for $f(z):=\sum_{k=1}^{\infty}e^{-\lambda_{k}z}c_{k}$ and some function F

$$\frac{f(z)-f(0)}{z}\to F(iIm(z)),$$

uniformly or in L1; as $Re(z)\to 0$ and $Im(z)\in [-\lambda,\lambda]$, where in our case $\lambda=\infty$. So assuming we can show convergence, the limit is

$$\frac{\sum_{k=1}^{\infty}c_{k}(e^{-\lambda_{k}iy}-1)}{iy}$$

and since $\sum a_{k}<\infty$ we ask that $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy}$ is well-defined. However, I think we can concoct $c_k$ s.t. $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy_{0}}=\infty$ for some fixed $y_{0}$.