For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple. Fix such a metric and denote the coresponding eigenvalues by

$$ \lambda_1, \lambda_2,\cdots $$

Using Weyl's asymptotic expansion we conclude that

$$\lambda_n\sim const . n^{m/2}. $$

Thus for any polynomial $P$ of degree $d>m/2$ we have

$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$

For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple. Fix such a metric and denote the coresponding eigenvalues by

$$ \lambda_1, \lambda_2,\cdots $$

Using Weyl's asymptotic expansion we conclude that

$$\lambda_n\sim \text{const}\cdot n^{m/2}. $$

Thus for any polynomial $P$ of degree $d>m/2$ we have

$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$

For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple. Fix such a metric and denote the coresponding eigenvalues by

$$ \lambda_1, \lambda_2,\cdots $$

Using Weyl's asymptotic expansion we conclude that

$$\lambda_n\sim const . n^{2/m}. $$

Thus for any polynomial $P$ of degree $>1$ we have

$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$

For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple. Fix such a metric and denote the coresponding eigenvalues by

$$ \lambda_1, \lambda_2,\cdots $$

Using Weyl's asymptotic expansion we conclude that

$$\lambda_n\sim const . n^{m/2}. $$

Thus for any polynomial $P$ of degree $d>m/2$ we have

$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$