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I'm currently interested in proving this theorem which I have been thinking for quite a while:

Define a Dirichlet Series $$\sum_{k=1}^{\infty}a_k e^{-\lambda_k z}$$ where $\lambda_k$ is a strictly increasing sequence and $z \in \mathbb{C}$.

I'm interested in proving that if the abscissa of convergence $\sigma_c <0$, then $$\sigma_c = \eta := \limsup_{n \to \infty} \frac{\log |R_n|}{\lambda_n}$$ where $R_n= \sum_{k=n+1}^{\infty}a_k$.

Anyone has any idea on how to start? I'm pretty much stuck after trying to use Abel's transformation like in the proof for when $\sum a_k$ is divergent.

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  • $\begingroup$ This is basically proved as Theorem 7 on page 13 here. There $\lambda_n = n$, but the exact same proof goes through in the general case. $\endgroup$ Nov 10 '14 at 5:10
  • $\begingroup$ Actually, looking at it further, a little work might be required in the second part of the proof (that $\sigma_c\leq \eta$) to pass to the general case. $\endgroup$ Nov 10 '14 at 5:42
  • $\begingroup$ @NickStrehlke I've successfully generalized the case for $\sigma_c \geq \eta$, but not for $\sigma_c \leq \eta$, and I realize there's something wrong in his proof. But anyway, thanks for the remark but I've already considered it and in fact, am still trying to work on it. $\endgroup$ Nov 10 '14 at 8:57
  • $\begingroup$ Right, my idea for generalizing the first part was essentially what you wrote (+1). I'd definitely be interested if you generalize the second part successfully! Maybe I'll work on it later or see if I can dig up a proof. $\endgroup$ Nov 10 '14 at 16:55
  • $\begingroup$ I found a proof that easily generalizes the reverse inequality. The reference is The General Theory of Dirichlet's Series by Hardy & Riesz. The argument I have in mind is on page 7 (look at (i)); it is actually directed toward proving an analogous formula for $\sigma_c$ in case it is positive, but I've worked through it and it can be adapted (it's very similar to the more specific argument given in the other reference). If you don't have access to the book I can add an answer later. $\endgroup$ Nov 10 '14 at 18:55
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I apologize for the delay in writing this up. Here is a proof of the remaining inequality (the one not proved in your answer, ireallydonknow). As I mentioned in a comment, this proof is a simple application of ideas found on page 7 of Hardy & M. Riesz's The General Theory of Dirichlet's Series. Here are a few remarks (mostly paraphrasing that book) to clarify the import of these formulas for other people who stumble across this question and wonder about its motivation. Basically, this formula is an analogue of the root test for the convergence of a power series (in fact, in the case $\lambda_n = n$ the two tests coincide). The situation is much more subtle in the context of Dirichlet series, however—a power series converges absolutely or diverges except possibly on a subset of a single circle, where it may converge conditionally (I think though that that subset can be quite complicated); a Dirichlet series, on the other hand, may converge conditionally (and not absolutely) on a strip or even on the entire plane. Hardy and Riesz, for instance, give as an example on page 9 the series $\Sigma (-1)^n (\log n)^{-s}/\sqrt{n}$.

With that, let me turn to the proof I mentioned in the comments. Right off the bat let me say that your formula is slightly off as written, in that the $\lambda_n$ appearing in the definition of $\eta$ should really be $\lambda_{n+1}$. To be completely precise, assume that $\sigma_c\leq 0$ and put \begin{align*} R_n = \sum_{k = n+1}^\infty a_k, \quad \eta = \limsup_{n\to\infty} {\log |R_n|\over \lambda_{n+1}};\tag{1} \end{align*} then $\eta = \sigma_c$. You'll see that the proof given in your own answer, ireallydonknow, carries through perfectly well with $\eta$ defined as in $(1)$, and shows that under these circumstances $\sigma_c\geq \eta$. The purpose of this answer, then, is to prove the reverse inequality $\sigma_c\leq \eta$. This reverse inequality will be proved if it can be shown that the Dirichlet series $\Sigma a_k e^{-\lambda_k s}$ converges whenever $s$ is a real number strictly larger than $\eta$, and that is what we'll show.

Thus, suppose $s>\eta$; we may assume moreover that $s<0$, since by hypothesis $\sigma_c\leq 0$. A tautological implication of the definition $(1)$ of $\eta$ is that $\log{|R_n|} \leq \lambda_{n+1}(\eta + o(1))$ as $n\to\infty$, and so $|R_n| \leq e^{\lambda_{n+1}(\eta + o(1))}$. Now use summation by parts to write (I'm denoting by $A$ the sum $\Sigma a_k$) \begin{align*} \sum_{k=1}^n a_k e^{-\lambda_k s} &= \sum_{k = 1}^n (R_{k-1} - R_k) e^{-\lambda_k s} \\ &= Ae^{-\lambda_1 s} - R_n e^{-\lambda_n s} + \sum_{k=2}^{n-1} R_k \left( e^{-\lambda_{k+1} s} - e^{-\lambda_k s}\right). \end{align*} First, $R_n e^{-\lambda_n s} \to 0$ as $n\to \infty$: \begin{align*} \log{|R_n|} - \lambda_n s < \log{|R_n|} - \lambda_{n+1} s \leq \lambda_{n+1}(\eta + o(1) - s) \to -\infty \quad \text{as $n\to\infty$,} \end{align*} using $\eta - s<0$ and $\lambda_{n+1}\to\infty$ for the limit relation at the end. The first inequality relies on $s<0$ and $\lambda_n<\lambda_{n+1}$ to conclude $-\lambda_n s<-\lambda_{n+1} s$.

We're now reduced to showing that $\Sigma R_k (e^{-\lambda_{k+1} s} - e^{-\lambda_k s})$ converges. (This is the part where we actually need my definition $(1)$ of $\eta$ instead of yours.) The sum actually converges absolutely. To see this, we follow the example of Hardy and Riesz and bound the general term as follows: \begin{align*} |R_k| (e^{-\lambda_{k+1} s} - e^{-\lambda_k s}) &\leq e^{\lambda_{k+1}(\eta + o(1))} (e^{-\lambda_{k+1} s} - e^{-\lambda_k s}) \\ & \leq |s| (\lambda_{k+1} - \lambda_k) e^{\lambda_{k+1}(\eta - s + o(1))}\\ &\leq |s| \int_{\lambda_k}^{\lambda_{k+1}} e^{x(\eta - s + o(1))}\,dx \end{align*} provided $k$ is large enough that $\eta - s + o(1) < 0$. Note that we use my definition $(1)$ of $\eta$ to get the third inequality, because we rely on $\lambda_{k+1}$ in the exponential factor arising in the bound on $|R_k|$; the definition in the question would lead to $\lambda_k$ instead, which isn't enough to make the third inequality hold. Anyway, the integrals are summable, and the proof is finished.

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  • $\begingroup$ So it is indeed what I thought it was... there is a stark difference between $\lambda_n$ and $\lambda_{n+1}$ in the formula. Thank you so much for the help! $\endgroup$ Nov 27 '14 at 6:26
  • $\begingroup$ Actually, I just realize I could do it with $\lambda_n$... just by modifying your proof... $\endgroup$ Nov 27 '14 at 8:01
  • $\begingroup$ @ireallydonknow I think this example rules out the $\lambda_n$ definition, but let me know if it looks like I've made an error somewhere: Take $a_k = e^{-k!} - e^{-(k+1)!}$ and $\lambda_k = k!$. Then $R_n = e^{-(n+1)!}$ and the definition of $\eta$ with $\lambda_{n+1}$ gives $\eta = -1$ while the definition with $\lambda_n$ gives $\eta = -\infty$. The Dirichlet series $\Sigma (e^{-k!} - e^{-(k+1)!})e^{-k!s}$ however doesn't converge for all negative $s$. $\endgroup$ Nov 27 '14 at 19:21
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@NickStrehlke This is the part for $\sigma_c \geq \eta$.

Let $s\leq 0$ be fixed so that the series converges at $s$. Define $B_n=\sum_{k=1}^{n}a_k e^{-\lambda_k s}$. We have $a_k e^{-\lambda_k s}=B_k - B_{k-1}$. Note that since the series converges at $s$, the partial sums of $B_n$ converge. Hence, the partial sums of $B_n$ are bounded, i.e., there exists some constant $M>0$ such that $|B_n|\leq M$ for all $n$. Now,

\begin{align} |R_n| \notag &=\left|\sum_{k=n+1}^{\infty} a_k e^{-\lambda_k s} \cdot e^{\lambda_k s} \right| \\ \notag &=\left|\sum_{k=n+1}^{\infty} [B_k - B_{k-1}] e^{\lambda_k s} \right| \\ \notag &= |[B_{n+1}-B_n]e^{\lambda_{n+1} s}+[B_{n+2}-B_{n+1}]e^{\lambda_{n+2} s}+...| \\ \notag &= |-B_n e^{\lambda{n+1} s}+\sum_{k=n+1}^{\infty} B_k (e^{\lambda_k s}-e^{\lambda_{k+1} s})| \\ \notag &\leq |B_n e^{\lambda_{n+1} s}|+\sum_{k=n+1}^{\infty} |B_k|(e^{\lambda_k s}-e^{\lambda_{k+1} s}) \\ \notag &\leq Me^{\lambda_{n+1} s}+M\sum_{k=n+1}^{\infty}(e^{\lambda_k s}-e^{\lambda_{k+1} s}) \end{align} If $s=0$, then $$\frac{\log |R_n|}{\lambda_n} \leq \frac{\log M}{\lambda_n}.$$ Taking the limit superior on both sides of the inequality, we get $$\eta=\limsup_{n \to \infty} \frac{\log |R_n|}{\lambda_n} \leq \limsup_{n \to \infty} \frac{\log M}{\lambda_n}=0=s.$$ If $s<0$, then $$\sum_{k=n+1}^{\infty}(e^{\lambda_k s}-e^{\lambda_{k+1} s})=e^{\lambda_{n+1}s},$$ and because $(\lambda_n)$ is strictly increasing, $$|R_n|\leq 2Me^{\lambda_{n+1}s} < 2Me^{\lambda_n s}.$$ Then we have $$\eta=\limsup_{n \to \infty} \frac{\log |R_n|}{\lambda_n} \leq \limsup_{n \to \infty} \frac{\log 2M}{\lambda_n}+s=s,$$ which implies that $\eta \leq s$. Since $\sigma_c = \inf s$, $\eta \leq \sigma_c$.

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