Let $(M^n,g)$ be a closed $n$ dimensional Riemannian manifold with $\mathrm{Ric}\_g\ge -K$, $(K\ge 0)$. Weyl's law(along with Karamata Tauberian Theorem) asserts that the eigenvalue $\lambda\_i$ of $-\Delta$ has the following asymptotic behavior
$$
\lambda\_i \sim c\_n\left(\frac{i}{\mathrm{Vol}\_g(M^n)}\right)^{2/n}\quad (i\to \infty)
$$
Is there a finer version of Weyl's law so that for $i$ large enough, the following uniform bound is true?
$$
\left|\lambda\_i-c\_n\left(\frac{i}{\mathrm{Vol}\_g(M^n)}\right)^{2/n}\right|\le C(K,n,D,\mathrm{Vol}\_g(M^n))
$$
Here $C(K,n,D, \mathrm{Vol}\_g(M^n))$ is a constant depending on the Ricci lower bound $-K$, dimension $n$, diameter $D$ of $M^n$ and the volume $\mathrm{Vol}\_g(M^n))$, the point is that it does not depend on $i$. If the conditions given at the beginning are not sufficient, what are the other conditions one can put on $M^n$ for the uniform estimates to hold (or any example to show it never holds)?


Thanks in advance.



A better version of Weyl's Law or uniform estimates of Laplacian higher eigenvalues
It seems unlikely that such a bound holds except in very special cases. For instance, it fails for round spheres, which have very large multiplicity of eigenvalues. In fact, for spheres the eigenvalue counting function has jumps of order $\lambda^{n-1}$.
On a round 2-sphere, for a constant eigenvalue $\lambda $, the second term in your estimate will vary on order $\lambda$, which makes a uniform bound on the left hand side impossible.


For generic metrics where closed geodesics are "sparse," it's possible to strengthen the standard Weyl law, but the refinement isn't strong enough to give uniform bounds on the difference between eigenvalues and the asymptotic formula from Weyl's law. I believe the state of the art refinement of Weyl's law is due to [Canzani and Galkowski](https://arxiv.org/abs/1903.08461), and their paper is a good reference. The relevant result is Theorem 7 on page 12.


