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I'm having trouble to evaluate a limit (shame on me!). It has to do with Weyl's asymptotic formula. It goes roughly as follows: $\Omega \subseteq M^n$. So the eigenvalues of the dirichlet laplacian satisfy

\begin{equation} \lim_{l \to \infty} \frac{\lambda_l}{l^{\frac{2}{n}}} = \frac{4\pi^2}{(\omega_n |\Omega|)^{\frac{2}{n}}} \end{equation}

Now I should be able to conclude that

\begin{equation} \lim_{l \to \infty} \frac{\frac{1}{l}\sum\limits_{i=1}^l\lambda_i}{l^{\frac{2}{n}}} = \frac{n}{n+2}\frac{4\pi^2}{(\omega_n |\Omega|)^{\frac{2}{n}}} \end{equation}

Thanks in advance

Bohrer
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1 Answers1

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You want $$\lim \frac{\sum_{i=1}^{l}\lambda _i}{l^{1+2/n}}.$$ Use discrete L'Hospital.

Compute instead the limit of the quotients of the discrete derivatives

\begin{align}\lim \frac{\lambda_l}{(l+1)^{1+2/n}-l^{1+2/n}}&=\lim \frac{\lambda_l}{l^{2/n}}\lim \frac{l^{2/n}}{(l+1)^{1+2/n}-l^{1+2/n}}\\&=\lim \frac{\lambda_l}{l^{2/n}}\lim \frac{1}{(l+1)(1+1/l)^{2/n}-l}\\&=\lim \frac{\lambda_l}{l^{2/n}}\lim \frac{1}{\sum_k\binom{2/n}{k}l^{-k}+\sum_k\binom{2/n}{k}l^{-k+1}-l}\\&=\lim \frac{\lambda_l}{l^{2/n}}\lim\frac{1}{1+\binom{2/n}{1}}\\&=\lim \frac{\lambda_l}{l^{2/n}}\frac{n}{n+2}\end{align}

Since this limit exists (because the given limit exists) then that's the value of the limit you were looking for.

user463383
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