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Consider a Dirichlet series $ \sum_n \frac{1}{n^s} $. At $ s = \sigma_c = 1 $ this series diverges to $ + \infty $ and it similarly diverges to $ +\infty $ for all $ s = \sigma < \sigma_c $.

On the other hand, a Dirichlet series $ \sum_n \frac{-1}{n^s} $. At $ s= \sigma_c = 1 $ this series diverges to $ - \infty $ and it similarly diverges to $ - \infty $ for all $ s = \sigma < \sigma_c $.

What would happen to a Dirichlet series where $ a_n $ takes both positive and negative values that vary to the left of abscissa of $ \sigma_c $? E.g., consider Dirichlet series $ \sum_n \frac{a_n\cos(\log n) }{n^s} $. Assume, at $ s = \sigma_c $ the series diverges to $ -\infty $. Would it similarly diverges to $ -\infty $ for all $ s = \sigma < \sigma_c $ or could it converge based on values $ a_n $.

I think the above is true considering the argument in the fundamental theorem (see page 3 at this link): if the series is converge the at $ s = \sigma_0 + it_0 $ it is convergent for all $ \Re(s) > \sigma_0 $. But somehow don't feel comfortable with my understanding.

To put my question in other words: if a Dirichlet series diverges to $ -\infty $ at $ s = \sigma_0 + it_0 $, will it diverge to $ -\infty $ for all $ s = \sigma + it_0 $ where $ \sigma < \sigma_0 $, irrespective of values of $ a_n $?

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A Dirichlet series cannot converge for any $\Re{(s)} = \sigma < \sigma_c$, but it need not diverge to $\pm\infty$ as can be shown by the Dirichlet Eta function, $\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}$. In this case $\sigma_c = 0$ (by the Alternating Series Test) and at $0$ the series diverges by oscillation. Even when the coefficients are all positive the series need not diverge at the abscissa of convergence as pointed out in this question.

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  • $\begingroup$ Thanks! If the series does diverge to $ +\infty $ or $ -\infty $ as shown in the first two examples, would it stay diverged in that direction? In other words, if it doesn’t oscillate, and diverges to one or the other infinity, is it monotonically increasing or decreasing? Or could it have a pole at $ \sigma_c$ and be convergent for $ \sigma_c -\epsilon $? $\endgroup$
    – Shree
    Commented Apr 29, 2018 at 2:36
  • $\begingroup$ @Shree By Theorem 1 on page 3 of the link you referenced above if a Dirichlet series converges for $s=\sigma + it$ it converges for any value of $s$ with real part greater than $\sigma$, hence a Dirichlet series cannot be convergent for any $\sigma_c-\epsilon$ If as in your examples the coefficients are all the same sign, then the series must diverge to plus or minus infinity for $\sigma < \sigma_c$ $\endgroup$
    – sharding4
    Commented Apr 29, 2018 at 3:27
  • $\begingroup$ are you implying- if the signs of $ a_n $ are not the same (like in $ \cos(\log n) $) we cannot have divergence to $ -\infty $ of course it must oscillate between plus and minus $ \infty$ or not have a single limit? Perhaps because in this case $ \lim{n\rightarrow\infty} a_n \neq 0 $ $\endgroup$
    – Shree
    Commented Apr 29, 2018 at 3:50
  • $\begingroup$ BTW, $ \cos (t \log n) $ is bounded, and $ \frac{1}{n^s} $ is decreasing. But $ \cos (t \log n) $ can change drastically to any value between -1 and + 1 from term to term, not promising alternate signs consistently. $\endgroup$
    – Shree
    Commented Apr 29, 2018 at 4:01
  • $\begingroup$ im interested in this question too. was it resolved? $\endgroup$
    – Penelope
    Commented May 28, 2021 at 3:48

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