Suppose $M$ is a compact Riemannian manifold of dimension $d$ without boundary, with Laplace Beltrami operator $\triangle$. We know that the spectrum of $\triangle$ (defined appropriately as a self - adjoint operator on $L^2(M)$) is discrete and consists of real numbers $\lambda_n$ that tend to $+ \infty$ as $n \to \infty$. The heat trace is then defined to be the sum $$ Z(t) = \sum_n e^{-\lambda_n t} \qquad (t > 0) $$ I am trying to understand how one can show directly that the sum converges. One thing that I think might help is Weyl's law which tells us something about the growth behaviour of the eigenvalues. Concretely we have $$ \lambda_n \backsim n^{2/d} \quad (n \to \infty) $$ so I think that once I understand the convergence of the series $$ \sum a^{n^{2/d}} $$ where $|a| < 1$, I have also understood the convergence of $Z(t)$. Here I am stuck -- basically the geometric series does not work as a model case (I think), and I also wonder whether this approach is the right to take ...
Thanks for your feedback and comments! I am destined to think it through myself, just need some help to progress in the right direction.
You can deduce the convergence of the series $\sum_n a^{n^{2/d}}$ with $0 < a < 1$ by first applying Cauchy's condensation test. Thus the original series converges iff the series $\sum_n 2^n a^{2^{2n/d}}$ converges.
Now apply the root test: This latter series converges if $$ \limsup_n 2 \cdot a^{2^{2n/d}/n} < 1 \, . $$ Since $2^{2n/d}/n = 2^{2n/d - \log_2 n} \to \infty$ as $n \to \infty$, this $\limsup$ is zero and convergence follows.