Given two increasing divergent sequences $a_n, b_n$ such that $\sum_{i=1}^\infty e^{-ta_i}=\sum_{i=1}^\infty e^{-tb_i}$ is it true that $a_n=b_n$?

Question

Given two strictly increasing and divergent sequences $a_n$ and $b_n$ such that the series $\sum_{i=1}^\infty e^{-ta_i}$ and $\sum_{i=1}^\infty e^{-tb_i}$ converge and are equal for every $t>0$, is it true that $a_n=b_n$ for every $n$?

This problem arised while I was studying some spectral geometry (and more specifically in an exercise where I'm required to prove that two compact riemannian manifolds with the same heat trace are isospectral). I tried to remove all of the details that seemed unnecessary to make this problem accessible to a broader audience, so there is some possibility that the statement of the problem is wrong.

I would be happy also with a counterexample.

Answer 1

Hint:

If you have an positive increasing divergent sequence $\left(c_n\right)$ such that $\sum e^{-c_nt}$ converges for all $t > 0$,

$$\lim_{t\to \infty} \sum_{n}e^{-c_kt} = 0$$

Let $g(t) = \sum_n e^{-c_nt}$ this function in a non increasing function so it is enough to prove that $g(m)\to 0$ for integers $m$. Or you can see that $g(m+1) \le e^{-c_1} g(m)$ So by induction, $g(m) \le g(1)e^{-c_1(m-1)} \to 0$

Answer 2

We'll prove by induction on $k\geq 0$ that $a_k=b_k$. Both the basis of the induction and the inductive step are proved in the same way so let's prove the inductive step. Let's suppose that $a_i=b_i$ for every $i\leq k-1$, therefore $$\sum_{j=k}^\infty e^{-a_j t}=\sum_{j=k}^\infty e^{-b_j t} \ \ \ \text{for every $t>0$. }$$ This implies that: $$\lim_{t\to \infty} \frac{\sum\limits_{j=k}^\infty e^{-a_j t}}{\sum\limits_{j=k}^\infty e^{-b_j t}}=1 $$ and since the sequences $(a_j)_{j\geq 0}$ and $(b_j)_{j\geq 0}$ are strictly increasing, we have $$\lim_{t\to \infty} \frac{e^{-a_k t}}{ e^{-b_k t}}=1 $$ and this concludes the proof.