# Prove that the Dirichlet eta function is monotonic

Let us consider $\eta(p):= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n^p}$ for $p>0$. Has anyone come along with an elementary proof that $\eta(x)$ is monotonically increasing on this set? By elementary I mean that this is a problem of real analysis, and there must be a solution without using analytic continuation for $\zeta(p)$ and its relations to $\eta(p)$. Another notable remark is that function $g(p):=\frac{1}{a^p}-\frac{1}{(a+1)^p}$ ($a\geq1$) can be both increasing or decreasing at $p=p_0$, depending on $a$, which makes this problem more complex than it seems at first glance. Thanks in advance!

• Where can we find a (non-elementary) proof of the mentioned statement? – GH from MO Sep 12 '14 at 21:53
• There is a proof here, based on an earlier result of Wang (which is unfortunately in Chinese): ajmaa.org/RGMIA/papers/v17/v17a103.pdf – GH from MO Sep 12 '14 at 22:29
• First they obtain logarithmic concavity... Thanks! Now I see this statement can hardly have a simple proof... – Alexander Sep 12 '14 at 23:11
• There is an easy proof that $\eta(p) \geq \eta(0) = 1/2$. We have $\eta(p) = (1/2) + \sum_{n=0}^{\infty} (1/2) (2n)^{-p} - (2n+1)^{-p} + (1/2) (2n+2)^{-p} \geq (1/2) \sum_{n=0}^{\infty} \sqrt{(2n)(2n+2)}^{-p} - (2n+1)^{-p} \geq 1/2$ where the first inequality is AMGM. I've been trying to adapt this to show $\eta'(p)>0$, but no luck yet. – David E Speyer Sep 13 '14 at 16:54
• @Alexander You can group terms of a conditionally convergent sum, you just can't rearrange them. Write out the partial sums of $(1/2) + 1/2 \sum (1/(2k-1)^s - 2/(2k)^s + 1/(2k+1)^s)$ and check that they approach the same limit as the partial sums of $\sum (-1)^{n-1}/n^s$. – David E Speyer Sep 15 '14 at 17:58

Theorem Let $f$ and $g$ be continuous rapidly decaying positive functions on $[0, \infty)$. Define $$F(s) = \int_{x=0}^{\infty} f(x) x^s \frac{dx}{x} \quad G(s) = \int_{x=0}^{\infty} g(x) x^s \frac{dx}{x}$$ Suppose that $f(x)/g(x)$ is increasing. Then $F(s)/G(s)$ is increasing.

Intuitively, as $s$ grows, the part with $x$ large contributes more to the integral, and $f/g$ is larger when $x$ is large.

Proof We want to show that $\frac{d}{ds} (F/G) = (F' G - F G')/G^2>0$. Differentiating under the integral sign, $$F'(s) G(s) - F(s) G'(s) = \int_{x=0}^{\infty} \int_{y=0}^{\infty} f(x) g(y) (\log x - \log y) x^s y^s \frac{dx \ dy}{xy}$$ $$= \int_{0 \leq x \leq y < \infty} f(x) g(y) (\log x - \log y) x^s y^s \frac{dx \ dy}{xy} + \int_{0 \leq y \leq x < \infty} f(x) g(y) (\log x - \log y) x^s y^s \frac{dx \ dy}{xy}$$ $$=\int_{0 \leq y \leq x < \infty} (f(x) g(y) - f(y) g(x)) (\log x - \log y) x^s y^s \frac{dx \ dy}{xy}. \quad (\ast)$$ (At the first line break, we split the domain of integration in two. We then interchange the variables $x$ and $y$ in the first integral and recombine the domains.)

For $x \geq y$ we have $\log x - \log y \geq 0$. Also, since $f/g$ is increasing, we have $f(x) g(y) - f(y) g(x) \geq 0$. So the integrand in $(\ast)$ is nonnegative, and so is the integral. $\square$

Now, take $f(x) = e^{-x}/(1+e^{-x})$ and $g(x) = e^{-x}$. It is obvious that $f(x)/g(x) = 1/(1+e^{-x})$ is increasing. Then $$G(s) = \int_{x=0}^{\infty} e^{-x} x^s \frac{dx}{x} = \Gamma(s)$$ and $$F(s) = \sum_{n=1}^{\infty} (-1)^{n-1} \int_{x=0}^{\infty} e^{-nx} x^s \frac{dx}{x} = \sum_{n=1}^{\infty} (-1)^{n-1} \Gamma(s) n^{-s} = \Gamma(s) \eta(s)$$ so $\eta(s)$ is increasing, as desired.

REMARK I didn't really need to differentiate. Let $s>t$, then rearranging $F(s)/G(s) \geq F(t)/G(t)$ in the same way leads to $$\int_{0 \leq y \leq x < \infty} \left( x^s y^t - x^t y^s \right) \left( f(x) g(y) - f(y) g(x) \right) \frac{dx \ dy}{xy} \geq 0$$ which is true for the same reasons. Probably the best formulation is that if $f(x)/g(x)$ and $s(x)/t(x)$ are increasing, than $$\left( \int f(x) s(x) dx \right) \left( \int g(x) t(x) dx \right) \geq \left( \int f(x) t(x) dx \right) \left( \int g(x) s(x) dx \right).$$

• Nice and useful! – GH from MO Sep 13 '14 at 19:47
• Your last inequality reminds me of en.wikipedia.org/wiki/Rearrangement_inequality – GH from MO Sep 14 '14 at 22:09
• @GHfromMO Agreed, rearrangement is the case $g(x)=1$, $s(x) = t(-x)$ (or $s(x) = t(x^{-1})$ if we are using $dx/x$). Chebyshev's inequality is $g(x) = t(x) = 1$. – David E Speyer Sep 14 '14 at 23:24
• A slightly more general version of your last inequality is the following. Let $I\subseteq\mathbb{R}$ be an interval endowed with a finite measure $dm$. If $a,b:I\to\mathbb{R}$ are increasing functions, then $(\int ab\,dm)(\int 1\,dm)\geq (\int a\,dm)(\int b\,dm)$. This, of course, is a variant of Chebysev's sum inequality, and the proof is a variant of your argument. Your inequality follows by putting $dm:=g(x)t(x)dx$, $a(x):=f(x)/g(x)$, $b(x):=s(x)/t(x)$. – GH from MO Sep 15 '14 at 5:43
• @Alexander Agreed. I think what I wrote is at a good level for communicating with other professionals but you are right, there are subtleties here, since $\sum (-1)^{n-1} e^{-nx}$ isn't uniformly convergent on $[0, \infty)$. If I were presenting this to an underaged analysis class, there would have to be a discussion about cutting the integral into $[a, \infty)$, where the convergence is uniform so we can interchange $\sum$ and $\int$, and $[0,a)$, where the integral is at most $\int_0^a x^s \frac{dx}{x} = a^s/s$. Then send $a \to 0$ to complete the computation. – David E Speyer Sep 15 '14 at 17:55

This is Theorem 3 (p. 10) in the Report:

J. van de Lune, Some inequalities involving Riemann's zeta-function, CWI Report ZW 50/75, CentruM Wiskunde & Informatica, Amsterdam,1975,

in which the proof does not use complex variables. The pdf of this Report can be found in the CWI repository

https://repository.cwi.nl/noauth/search/fullrecord.php?publnr=6895

• Thanks for the link! But in fact they use equality $\eta(s)=(1-2^{1-s})\zeta(s)$, and for $s<1$ this involves analytic continuation for ζ(s). – Alexander Sep 15 '14 at 12:45