Suppose $s$ is a complex number with $\Re(s) \in (0,1]$ and $\{a\_n\}$ is a complex sequence converging to $a \neq 0$. Must the Dirichlet series $$\sum\_{n=1}^\infty\frac{a\_n}{n^s}$$ diverge?


I asked this question on math.stackexchange on 2011-12-13. Here is the link: <https://math.stackexchange.com/questions/91218/divergence-of-dirichlet-series>.


For reference, put $s = r + it$. It is easy to see that the above series must converge if $r > 1$ and must diverge if $r \leq 0$ or if $s = 1$. On math.stackexchange I was able to prove divergence for $0 < r < 1$, using a result from Knopp. This leaves the question open for $s = 1 + it$, where $t \neq 0$.



Divergence of Dirichlet series
The answer to your question is yes, the series must always diverge when $\Re (s)\in (0,1]$. The proof is a little delicate but here is the argument:


First of all to reiterate what you already mentioned, since the $a\_n$'s have a non-zero limit, the abscissa of absolute convergence and the abscissa of convergence are both equal to $1$. Also the series clearly does not converge absolutely for $\Re (s)=1$, and it doesn't converge at all for $s=1$, so the only question is whether or not it converges conditionally for $s=1+it,~t\not= 0$.


Now for $t\not= 0$ let $$S\_N=\sum\_{n=1}^N\frac{a\_n}{n^{1+it}}.$$
By a simple power series argument, for all $K,N\in\mathbb{N}$ with $K\le N/2$ we have that
$$t(\log (N+K)-\log N)<\frac{ctK}{N},$$
for some universal constant $c>0$. Thus for $\epsilon >0$ if $N$ is large enough (depending on $t$) then choosing $$K=\left[\frac{\epsilon N}{ct}\right],$$
gives that $t(\log (N+K)-\log N)<\epsilon$.


Finally assume that $\epsilon$ is small and that $N$ is chosen as large as you need so that the quantities $a\_n\exp(-it\log n)$ are all close to some constant $c'$, with $|c'|\approx |a|$, for all $N< n\le N+K.$
Then since the argument of the summand can not change by more than $\epsilon$ over this range we have that
$$|S\_{N+K}-S\_N|=\left|\sum\_{n=N+1}^{N+K}\frac{a\_n\exp(-it\log n)}{n}\right|\gg |a|\sum\_{n=1}^K\frac{1}{N+n}.$$
Pulling out the $N$ from the denominator gives
$$|S\_{N+K}-S\_N|\gg \frac{|a|K}{N}\gg 1.$$
The final implied constant depends on $\epsilon$ and $t$, but both of these parameters are fixed (note that we don't have to let $\epsilon$ tend to $0$ to make this work, we just need it to be small). Therefore the sequence $S\_N$ is not a Cauchy sequence and the Dirichlet series is not conditionally convergent for any value of $t$.


The underlying reason why the series don't converge is that the phase of the summand is not increasing fast enough to cause the necessary amount of cancellation.


