How close to zero can a Dirichlet series get? Suppose I have an integral Dirichlet series $f(s) = \sum c_n n^{-s}$, $c_n \in \mathbb{Z}$, with at least one non-zero term $c_N$.  Suppose furthermore that this series converges absolutely and uniformly for $\mathrm{Re}(s) > 1 + \delta$ for any $\delta > 0$ (so that $c_n$ grows slower than $n^\epsilon$ for any $\epsilon$).
I want $f(s)$ to be "small" (in terms of $N$) for fixed $\mathrm{Re}(s)$ and a range of $\mathrm{Im}(s)$.  How small can I get it over any given range?
I think one can take sums of $(N + \Delta n)^{-s}$ for $\Delta n \approx O(\ln N)$ and cancel out terms in the Taylor series up to order $N^{-(s + O(\ln N))}$, all while keeping the coefficients polylog in N for $Im(s) \approx O(\ln N)$.  Can we do any better?  What about larger values of $s$?
 A: I alluded to one construction that gives small values for $f(s)$ in a given range of Im(s).  I will elaborate on that construction here.
Consider the Taylor expansion of $(N+x)^{-s}$.  This is 

$(N+x)^{-s} = N^{-s} * (1 - \frac{sx}{N} + \frac{s(s+1)x^2}{2N^2} - \ldots )$.

We are going to truncate this at some finite order $k$, and allow $x$ to take on different integer values.  We then obtain a number of terms polynomial in $s$ (times $N^{-s}$) and we want a linear combination of these terms to vanish for all $s$ -- that implies $x$ takes on at least $k$ different values.  
Define the $c_n$ to be the coefficients in this sum.  We would like the remaining error term in the Taylor series (the terms of order $s^{(k+1)}/N^{k+1}$ and higher) to be $o(N^{-k})$.  This implies that $\sum c_n (n-N)^{k+1} s^{(k+1)} / k!$ is $o(N)$.  Remember that we have at least $k-1$ nonzero terms in this sum, and so there is at least one $n$ with $|n-N| \geq (k-1)/2$.
To ensure that this sum is $o(N)$, we should take $k$ to be $o(\ln N)$ and similarly for $s$.  We can readily find a set of $n$ and $c_n$ which will keep the sum at $o(N)$ -- I believe the $k$th-order difference should work, although I am not certain.
