If $\lambda_n \sim \mu_n$, is it true that $\sum \exp(-\lambda_n x) \sim \sum \exp(-\mu_n x)$ as $x \to 0$?

I'll consider the case $\lambda_n\sim an$. I claim that $$\sum_{n}e^{-\lambda_n x} \sim \frac{1}{ax}$$ Given $\epsilon>0$, pick $N$ such that $|\lambda_n/n-a|<\epsilon$ for $n\ge N$. Estimate the tail $$\sum_{n\ge N}e^{-(a+\epsilon) n x}\le \sum_{n\ge N}e^{-\lambda_n x} \le \sum_{n\ge N}e^{-(a-\epsilon) n x}$$ Sum the geometric series and multiply the result by $x$: $$\frac{x\, e^{-(a+\epsilon) N x}}{1-e^{-(a+\epsilon) x}}\le x \sum_{n\ge N}e^{-\lambda_n x} \le \frac{x\,e^{-(a-\epsilon) N x}}{1-e^{-(a-\epsilon) x}}$$ As $x\to 0$, we the left-hand side tends to $(a+\epsilon)^{-1}$ while the right hand side tends to $(a-\epsilon)^{-1}$. Since the contribution of the terms with $n<N$ is $O(x)$, we have $$(a+\epsilon)^{-1}\le \liminf_{x\to 0} x\sum_{n}e^{-\lambda_n x} \le \limsup_{x\to 0} x\sum_{n}e^{-\lambda_n x} \le (a-\epsilon)^{-1}$$ And since $\epsilon$ was arbitrary, we are done.