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Let $(a_n)_{n \ge 1}$ be a sequence of non-negative real numbers such that $\sum_{n \ge 1} a_n = \infty$, and set $\lambda_n := a_1 + \cdots + a_n$ for each $n$. Then the (generalized Dirichlet) series $\sum_{n \ge 1} a_n e^{-\lambda_n s}$ converges for all $s > 0$, as its abscissa of convergence is $0$ by the main result of Chapter II, Section 6 from:

G. H. Hardy and M. Riesz, The General Theory of Dirichlet's Series, Cambridge Univ. Press: Cambridge, 1915.

So my question is:

Q. What about sufficient conditions for having that $\sum_{n \ge 1} a_n e^{-\lambda_n s} \sim \frac{1}{s}$ as $s \to 0^+$?

To be clear: I'm just looking for references.

For the record: I'm especially, though not uniquely, interested in the case where $a_n = n^{\alpha}$ for all $n$, with $\alpha$ being a given exponent $\ge -1$.

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Set $\lambda_0:=0$. We have $$ s^{-1}=\sum_{n=1}^{\infty} s^{-1}\left(e^{-\lambda_{n-1}s}-e^{-\lambda_n s}\right)=\sum e^{-\lambda_ns}\frac{e^{a_ns}-1}{s}\geqslant \sum a_ne^{-\lambda_ns}. $$ On the other hand, $$ s^{-1}=\sum_{n=1}^{\infty} s^{-1}\left(e^{-\lambda_{n-1}s}-e^{-\lambda_n s}\right)=\sum e^{-\lambda_{n-1}s}\frac{1-e^{-a_ns}}{s}\leqslant \sum a_ne^{-\lambda_{n-1}s}=a_1+\sum a_{n+1}e^{-\lambda_ns}. $$

Thus if $a_n\sim a_{n+1}$, we have desired relation $\sum a_ne^{-\lambda_ns}\sim s^{-1}$ for sure. Of course, this may be weakened.

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  • $\begingroup$ Thanks, and +1. However, let me note that I asked for a reference, not for a proof. The more general the case covered by the reference is, the happier I am. $\endgroup$
    – Salvo Tringali
    Commented Feb 11, 2016 at 14:11
  • $\begingroup$ Yes, I see. Unfortunately, I do not know references. $\endgroup$
    – Fedor Petrov
    Commented Feb 11, 2016 at 14:16

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