# 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 $x>0$.

I believe this is true (and some numerical examples agree), but I can't see how to prove it. Intuitively, if we fix $N$ large and split up the sum like

$$\underbrace{\sum_{n=1}^{N}}_{P} + \underbrace{\sum_{n=N+1}^{\infty}}_{Q}$$

Then $P$ is controlled by small $x$ and $Q$ is... I really don't know. Since $N$ is large we should be able to use the fact that $\lambda_n \sim \mu_n$ as $n \to +\infty$ here, or something like that.

At least for $P$ we have

$$\sum_{n=1}^{N} \exp(-\lambda_n x) = N - \left(\sum_{n=1}^{N} \lambda_n\right)x + O(x^2)$$

as $x \to 0$.

Any tips would be appreciated.

Edit.

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.

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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? –  ACARCHAU 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