User richard hevener - MathOverflow

Divergence of Dirichlet series

2011-12-22T15:24:27Z

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 $t \neq 0$.

Answer by Richard Hevener for Divergence of Dirichlet series

2011-12-29T23:53:23Z

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.

Answer by Richard Hevener for Divergence of the Dirichlet series for the Riemann zeta function for Re s = 1, Im s <> 0

2011-12-06T03:05:11Z

I wanted to correct a couple of the inequalities in David Speyer's answer and comments.

I believe one in the answer should read $\left| \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq |s|/(2n^{{Re}(s)+1})$. (This has now been corrected by Mr. Speyer, omitting the "2" in the denominator, which is fine.)

Also, I believe the one in the following comment should read $|f'(x) (x - \lfloor x \rfloor)| \leq |s|/|x|^{{Re}(s)+1}$.

Answer by Richard Hevener for If $f$ is infinitely differentiable then $f$ coincides with a polynomial

2011-07-31T02:39:00Z

In Andrey Gogolev's answer the following two assertions appear:

"It is clear that $X$ is a non-empty . . . set" and "Now consider any maximal interval $(c,e) \subset ((a,b) - X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$."

These are true, but perhaps not transparently obvious. In attempting to fill the gaps, I developed a variation of the proof which requires neither the observation that $X$ has no isolated points nor any argument about degrees of polynomials. Here is my adaptation, borrowed freely from Gogolev:

I use the symbol "$\bot$" for "contradiction."

Define $I = [0,1]$ and $X = \{x \in I: \forall (a,b) \ni x: f|_{(a,b) \cap I} \; is \; not \; a \; polynomial\}$ .

We first establish the following:

Lemma: Suppose $[c,d] \subset I$ is an interval on which $f$ coincides with a polynomial $p$. Then there exists a maximal subinterval $[cm,dm]$ having the properties $[c,d] \subset [cm,dm] \subset I$ and $f = p$ on $[cm,dm]$. Furthermore, $cm \in X \cup \{0\}$ and $dm \in X \cup \{1\}$.

Proof: Let $cm$ = LUB $\{x: f(x) \neq p(x)\} \cup \{0\}$ and $dm$ = GLB $\{x: f(x) \neq p(x)\} \cup \{1\}$. It is clear that $[cm,dm]$ is maximal. Supppose that $cm \not \in X$ and $cm \neq 0$. Then we can find another interval $(u,v)$ with $cm \in (u,v) \subset I$ on which $f$ coincides with a polynomial $q$. But on $[cm,v]$ we have $f = p = q$, whence $f = p$ on $[u,dm]$. Since $u < cm$, we see that $[cm,dm]$ is not maximal ($\bot$). Therefore, $cm \in X$ or $cm = 0$. Likewise, $dm \in X$ or $dm = 1$.

Now we begin the proof-by-$\bot$ of the main result. Suppose that $f$ is not a polynomial on $I$.

If $X = \emptyset$, we begin with any $[c,d]$, and the lemma tells us that $cm = 0$ and $dm = 1$, so $f$ is a polynomial on $I$ ($\bot$). Thus, $X \neq \emptyset$. Now define $S_n = \{x: f^{(n)}(x) = 0\}$. $X$ and $S_n$ are clearly closed. Applying the Baire category theorem to the covering ${X \cap S_n}$ of the complete metric space $X$, we get that there exists an interval $(a,b)$ such that $(a,b) \cap X \neq \emptyset$ and $(a,b) \cap X \subset S_n$ for some $n$. (It is important here that $S_n$ is closed.)

Put $J = (a,b) \cap I$, and let $a1$ and $b1$ be the left and right end-points of $J$. (Observe that it is possible that $a1 = 0$ or $b1 = 1$, so J may not be open.) If $J \subset S_n$, then $f$ is a polynomial on $J$, whence $(a,b) \cap X = (a,b) \cap I \cap X = J \cap X = \emptyset$ ($\bot$). Thus, we can choose a point $t \in J - S_n$. Now $t \not \in X$, since $(a,b) \cap X \subset S_n$. Therefore, we can find an interval $(c,d) \ni t$ such that $f$ coincides with a polynomial $p$ on $(c,d) \cap I$. Furthermore, $f = p$ on the closure of $(c,d) \cap I$, which is an interval of the form $[c1,d1] \subset I$. Apply the lemma to $[c1,d1]$ to obtain a maximal interval $[cm,dm]$ having the stated properties. Since $t \not \in S_n$ and considering $p$, we see that $cm \not \in S_n$. Suppose $cm > a1$. Then we have $a \le a1 < cm \le c1 \le t < b$, so $cm \in (a,b)$. From the lemma, $cm \in X$, since $cm > a1 \ge 0$. Thus, $cm \in (a,b) \cap X \subset S_n$ ($\bot$). Therefore, $cm \le a1$. Likewise, $dm \ge b1$. Thus, $f$ is a polynomial on $J \subset [a1,b1] \subset [cm,dm]$, whence, as above, $(a,b) \cap X = \emptyset$ ($\bot$). We are at last forced to conclude that $f$ must indeed be a polynomial on $I$.

Power function inequality

2011-10-19T17:32:03Z

Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ .

I recently discovered this result. I am sure it is known, but it is new to me. It is quite easy to prove if $p$ is an integer, even a negative one. I have a proof in the general case above, but it seems overly complicated. Can someone provide a simple demonstration?

Comment by Richard Hevener

2011-12-29T23:39:01Z

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.

Comment by Richard Hevener

2011-10-22T05:23:26Z

Thanks for the advice, Gerhard.

Comment by Richard Hevener

2011-10-20T19:12:49Z

Beautiful argument--thanks. As cardinal observed over at math.stackexchange, you don't need Jensen, just convexity.

Comment by Richard Hevener

2011-10-20T19:10:46Z

I wanted to explain that I noticed this result while studying an argument of Mazur's that Lp is homeomorphic to L1 for 1 < p < oo. It went unanswered for a week on math.stackexchange before I posted it here.