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 $\mathbb{R}^{2}$:


1. $\lambda\_{k}<\lambda\_{k+1}<....$
2. $\lambda\_{k}=\frac{4\pi}{|D|}k+c\_{2}\sqrt{k}+o(\sqrt{k})$
3. $\sum^{k}\_{j\geq 1}\lambda\_{j}\geq \frac{2\pi}{|D|}k^{2}\Rightarrow \lambda\_{k}\geq \frac{2\pi}{|D|}k$ (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 $y\_{0}$. 



Tauberian theorem $\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} $
The $\lambda\_k$ correspond to the Laplacian eigenvalues of a domain in $\mathbb{R}^2$ implies that
\begin{equation}
0<\lambda\_1\le \lambda\_2\le\cdot\cdot\cdot\le \lambda\_n\le\cdot\cdot\cdot
\end{equation}
and $\lambda\_n\rightarrow\infty $ as $n\rightarrow\infty$. Hence for any $t>0$
\begin{align}
\sum\_{n=1}^{\infty}e^{-\lambda\_nt}c\_n&=\sum\_{n=1}^{\infty}[C(n)-C(n-1)]e^{-\lambda\_nt}\\
&=\sum\_{n=1}^{\infty}C(n)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right]\\
&=C(\infty)+\sum\_{n=1}^{\infty}o\_n(1)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right],
\end{align}
by Abel transform, where
$$C(n)=\sum\_{k=1}^{n}c\_k ~\mbox{for}~ n\ge 1 $$
and 
$$o\_n(1)=\sum\_{k=1}^{\infty}c\_k-C(n)=C(\infty)-C(n)\rightarrow 0~as~ n\rightarrow \infty.$$
Then it is obviously that
$$\sum\_{n=1}^{\infty}o\_n(1)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right]\rightarrow 0~ as~t\rightarrow 0+.$$


In fact, for any $\varepsilon>0$, there exist an $N\_{\varepsilon}\in\mathbb{N}$ such that for all $n>N\_{\varepsilon}$, 
$|o\_n(1)|< \varepsilon$ 
holds. Furthermore,
$$\left|\sum\_{n=1}^{\infty}o\_n(1)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right]\right|\le \left|\sum\_{n=1}^{N\_{\varepsilon}}o\_n(1)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right]\right|+\epsilon\sum\_{n>N\_{\varepsilon}}\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right].$$
Namely,
$$\left|\sum\_{n=1}^{\infty}o\_n(1)\left[e^{-t\lambda\_n}-e^{-t\lambda\_{n+1}}\right]\right|\le \max\_{1\le k\le N\_{\varepsilon}}|o\_k(1)|e^{\lambda\_1t}+\epsilon.$$
The next let $t\rightarrow 0$ and then let $\varepsilon\rightarrow 0$, then the results is obvious. 


