# Divergence of Dirichlet series

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: http://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 $t \neq 0$.

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The results you mention seem to follow from math.nus.edu.sg/~chanhh/MA4263/Chapter6.pdf theorem 6.3.1 –  Igor Rivin Dec 22 '11 at 15:38
I added a small correction to your proof for $s=1$ on SE (it awaits "peer review" as I am not registered there). Your explanation that either the real or the imaginary part of $(a_n)$ is bounded away from zero was not sufficient, e.g. $\sum_n (-1)^n/n$ has this feature but it converges. You need that either the real or the imaginary part of $(a_n)$ tends to a nonzero real number. –  GH from MO Dec 23 '11 at 13:33
Yes, the theorem at math.nus.edu is much like the one I used from Knopp. Also, thanks, GH--I have clarified my argument on SE. –  Richard Hevener Dec 29 '11 at 23:39

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.

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I had a devil of a time figuring out the previous answer, but I now believe it to be fundamentally sound. Based on that answer, I am providing another one with a number of gaps filled in.

First note that we can assume WLOG that $a = 1$. Under this assumption we will show that either the real or the imaginary part of $\sum_{n=1}^\infty \frac {a_n} {n^s}$ diverges, where $s = 1 + it$ and $t \neq 0$. We also assume WLOG that $t > 0$. (Otherwise, replace $t$ by $-t$ in appropriate places below.) Put $a_n = u_n + iv_n$, where $u_n \rightarrow 1$ and $v_n \rightarrow 0$. Note that $$\Re(\frac {a_n} {n^s}) = (u_n \cos (t \log (n)) + v_n \sin (t \log (n))) / n$$ and $$\Im(\frac {a_n} {n^s})= (-u_n \sin (t \log (n)) + v_n \cos (t \log (n))) / n.$$

There must either be infinitely many values of n for which $|\cos (t \log (n))| > 1/2$ or infinitely many values of n for which $|\sin (t \log (n))| > 1/2$ . We assume the former. (Otherwise, switch the roles of sin and cos, and work with the imaginary part of the series instead of the real part; the necessary changes are fairly straightforward.) Likewise, WLOG we assume that there are inf. many values of n for which $\cos (t \log (n)) > 1/2$, as opposed to $< -1/2$. (Otherwise, replace cos by -cos and $\Re$ by $-\Re$ in appropriate places below.)

Now $\cos(x)$ is uniformly continuous on the real line. Select $\delta > 0$ such that if $|x - y| < \delta$, then $|\cos(x) - \cos(y)| < 1/4$ . We can also assume that $\delta < t$.

To prove divergence under the above assumptions, we will show that the sequence of partial sums of the real part of our series is not Cauchy. Let $M$ be a positive integer.

First find $M_1 > M$ such that for all $n > M_1$, $u_n > 1/2$ and $|v_n| < 1/16$. Choose $N$ an integer such that $N > \max(M_1, 2t / \delta)$ and $\cos (t \log (N)) > 1/2$ . Now $N \delta / 2t > 1$, whence $N \delta / t - N \delta / 2t > 1$. Thus, we can select $K$ an integer with $1 < N \delta / 2t < K < N \delta / t$ . Let $n$ be any integer with $n > N$. From calculus we know that $$t(\log (n) - \log (N)) < t(n - N) / N.$$
Observe that if $N < n \leq N + K$ , then $$t(\log (n) - \log (N)) < tK / N < \delta,$$ whence $$|\cos(t \log (n)) - \cos(t \log (N))| < 1/4.$$ Therefore, \eqalign{&u_n \cos (t \log (n)) + v_n \sin (t \log (n))\geq u_n \cos (t \log (n)) - |v_n \sin (t \log (n))|\cr&\qquad\gt \cos (t \log (n)) / 2 - 1/16\cr&\qquad\geq (\cos (t \log (N)) - |\cos (t \log (n)) - \cos (t \log (N))|) / 2 - 1/16\cr&\qquad\gt (1/2 - 1/4) / 2 - 1/16 = 1/16.\cr}

It follows that \eqalign{\sum_{n=N+1}^{N+K} \Re(\frac {a_n} {n^s}) &\gt \sum_{n=N+1}^{N+K} 1 / 16n \gt K / (16(N + K))= 1 / (16(N / K + 1))\cr& \gt 1 / (16(2t / \delta + 1))= \delta / (16(2t + \delta)) \gt \delta / 48t.\cr}
Thus, the sequence of partial sums of the real part of our series is not Cauchy.

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Richard, if you're still around, I've done rather a lot of reformatting of your answer. You might want to check to see whether I've messed anything up. –  Gerry Myerson Dec 6 '12 at 0:13