Consider $a_n$ a real valued sequence and define $D_{1,1,1}(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$ which converges in some half plane $\Re s =c.$ Define $D_{r,h,k}(s) = \sum_{n=1}^\infty \frac{e^{2\pi i \frac{hrn}{k}} a_n}{n^s}.$

Question: Suppose $D_{1,1,1}(s)$ has an analytic continuation to a half plane $R$ with a simple pole at $s=c.$ Is is necessiarly true that $D_{r,h,k}(s)$ must have an analytic continuation to $R$ with at most a simple pole at $s=c?$

EDIT: This question was answered by GH in the negative. His response leads me to ask a follow up question. Does the above statement hold if $a_n\geq 0?$

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    $\begingroup$ I fixed your tags from [number] and [theory] to [nt.number-theory] $\endgroup$ – David Roberts May 12 '11 at 2:58
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    $\begingroup$ Is there supposed to be an $n$ in the exponential? The $h$ and $r$ intervene only through their product, so why are they separate indices? Are you thinking about twisting Dirichlet series that arise in some interesting way - if so, some context would help. $\endgroup$ – Ramsey May 12 '11 at 3:33
  • $\begingroup$ Yes there is supposed to be a $n$ in the exponential. Unfortunately, I have trouble drafting things typo free. $\endgroup$ – Daniel Parry May 12 '11 at 4:00
  • $\begingroup$ @Daniel: I answered your question in the negative. $\endgroup$ – GH from MO May 12 '11 at 4:12
  • $\begingroup$ @GH: Thanks for the counterexample. Its such an interesting sequence to choose. How did you think of it? $\endgroup$ – Daniel Parry May 12 '11 at 4:18

A counterexample is provided by $c=1$ and $a_n=\sum_{d\mid n}(-1)^d$. Indeed, $$ D_{1,1,1}(s)=(2^{1-s}-1)\zeta(s)^2 $$ has a simple pole at $s=1$, but $$ D_{1,1,2}(s)=\sum_{n=1}^\infty \frac{(-1)^na_n}{n^s}=(1-2^{1-s}+2^{1-2s})\zeta(s)^2 $$ has a double pole at $s=1$.

EDIT: Daniel asked in a comment how I came up with this sequence, so I try to explain. Vaguely, the original question asks the following: can we create a pole of a Dirichlet series by additive twists? In other words, can the coefficients of a Dirichlet series obey much less cancellation along arithmetic progressions than originally? The answer to this vague reformulation is obviously yes, e.g. there is a lot of cancellation in the formal sum $1+(-1)+1+(-1)+\dots$ but there is no cancellation in its formal subsums $1+1+1+\dots$ and $(-1)+(-1)+(-1)+\dots$. So a natural candidate for creating a pole by additive twists is $\sum_n (-1)^n/n^s$ which can be twisted to $\sum_n 1/n^s$. Indeed, here we create a pole at $s=1$, because the first series equals $(2^{1-s}-1)\zeta(s)$ but the second series equals $\zeta(s)$. Now this does not answer the original question since $(2^{1-s}-1)\zeta(s)$ has no pole at all. We can remedy this by considering $(1-2^{1-s})\zeta(s)^2$ instead, i.e. by convolving the sequence $(-1)^n$ with $1$. This is my final example above: I simply checked that it worked, i.e. it creates a double pole from a single one!

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  • $\begingroup$ So I see that this depends on the fact that the sequences can be positive and negative. Would you know if this type of method would fail for $a_n$ all positive? $\endgroup$ – Daniel Parry May 12 '11 at 14:21
  • $\begingroup$ The question with $a_n\geq 0$ is certainly more difficult, I will think about it as time permits. In that case, by Landau's theorem, the original series converges for $\Re s>c$, so the same is true for the twisted series. In particular, any additional pole must lie in $\Re s\leq c$. I suspect that there is no counterexample which grows only polynomially slightly to the left of $c$, but I have no time to work on this now. $\endgroup$ – GH from MO May 13 '11 at 6:47

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