TITLE: If $\lambda_n \sim \mu_n$, is it true that $\sum \exp(-\lambda_n x) \sim \sum \exp(-\mu_n x)$ as $x \to 0$?
QUESTION [13 upvotes]: If $\lambda_n,\mu_n \in \mathbb{R}$, $\lambda_n \sim \mu_n$ as $n \to +\infty$, and $\mu_n \to +\infty$ as $n \to +\infty$, is it true that
  $$
\sum_{n=1}^{\infty} \exp(-\lambda_n x) \sim \sum_{n=1}^{\infty} \exp(-\mu_n x)
$$
  as $x \to 0^{+}$?
In other words, is it true that
  $$
\lim_{x \to 0^+} \frac{\sum_{n=1}^{\infty} \exp(-\lambda_n x)}{\sum_{n=1}^{\infty} \exp(-\mu_n x)} = 1?
$$

Note that since $\mu_n \to +\infty$ we must also have $\lambda_n \to +\infty$ to ensure that $\lambda_n \sim \mu_n$ as $n \to +\infty$, i.e. that
$$
\lim_{n \to +\infty} \frac{\lambda_n}{\mu_n} = 1.
$$
We also assume that each series converges for all $x>0$.

I believe this is true (and some numerical examples agree), but I can't see how to prove it.  Using the idea presented in this answer we have an upper bound like
$$
\sum_{n=1}^{\infty} e^{-\lambda n x} \leq \sum_{n=1}^{\infty} e^{-(1-\epsilon)\mu_n x} + O(1)
$$
with an analogous lower bound, where the term in $O(1)$ in bounded independently of $x$ (but does depend on $\epsilon$).  So, dividing through by $\sum_n e^{-\mu_n x}$, we're really interested in whether
$$
\lim_{\epsilon \to 0} \lim_{x \to 0^+} \frac{\sum_{n=1}^{\infty} e^{-(1-\epsilon)\mu_n x}}{\sum_{n=1}^{\infty} e^{-\mu_n x}} = 1.
$$
If this were true the result would follow.
Sometimes it's possible to show this a posteriori if we know an elementary closed form or asymptotic for $\sum_n \exp(-C\lambda_n x)$ valid for all $C$ in some neighborhood of $1$ and small $x > 0$, as was the case in the second half of this answer.  In this question I am interested in the case when we do not.

It was noted by PavelM in the comments that it may very well be false when $\lambda_n$ is almost $\log n$.
I am definitely interested in the general question.  However, I am specifically interested in the special case where
$$
\lambda_n \sim a n
$$
as $n \to \infty$ for some constant $a > 0$.  Any help with this specific problem would likewise be much appreciated.

REPLY [4 votes]: 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.