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

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.

• I have no idea what I'm talking about but I would try Taylor series of $exp$, then swap the two infinite sums and then see what happens. – xavierm02 Jan 13 '13 at 18:14
• For two sequences of positive numbers $a_n$ and $b_n$, if $a_n \sim b_n$ and either series $\sum_{n \geq 1} a_n$ or $\sum_{n \geq 1} b_n$ diverges (that is, tends to $\infty$) then $\sum_{n=1}^m a_n \sim \sum_{n=1}^m b_n$ as $m \rightarrow \infty$. This might be useful. You might need to distinguish between the sequences $\lambda_n$ and $\mu_n$ each tending to $0$ (like $1/n$) or to $\infty$ (like $n$). – KCd Jan 13 '13 at 18:14
• Are $\lambda_n, \mu_n > 0$ for all n? – Arin Chaudhuri Jan 13 '13 at 18:15
• @ACARCHAU they are eventually positive. I guess you can assume that they are always positive, if you must. – Antonio Vargas Jan 13 '13 at 18:26
• I'm inclined to think this is false. When $\lambda_n=\log n$, the series becomes $\sum n^{-x}$, which blows up already at $x=1$. Anything bigger, i.e., $\lambda_n/\log n\to \infty$, will produce a convergent series for all $x>0$. This suggests that the sum is very sensitive to the size of $\lambda_n$ when $\lambda_n$ is close to being logarithmic. I wonder about things like $\lambda_n=\log n\log\log n$ and $\mu_n = \lambda_n+\log n$... Numerical experiments are not reliable here: for a computer, $\log \log$ is a bounded function. – user53153 Jan 13 '13 at 21:47

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.
• Ah, thank you. This is very clear. A similar argument should work for $\lambda_n \sim n^k$, where $k > 0$ is an integer. Cheers! – Antonio Vargas Jan 13 '13 at 23:13
• @AntonioVargas Probably. The series $\sum e^{-n^k x}$ should be asymptotic to $\sum m^{k^{-1}-1} e^{-m x }$, which is a polylogarithm of negative order. – user53153 Jan 13 '13 at 23:23