24
$\begingroup$

Set $$ g(x)=\sum_{k=0}^{\infty}\frac{1}{x^{2k+1}+1} \quad \text{for} \quad x>1. $$

Is it true that $$ \frac{x^{2}+1}{x(x^{2}-1)}+\frac{g'(x)}{g(x)}>0 \quad \text{for}\quad x>1? $$ The answer seems to be positive. I spent several hours in proving this statement but I did not come up with anything reasonable. Maybe somebody else has (or will have) any bright idea?

Motivation? Nothing important, I was just playing around this question: A problem of potential theory arising in biology

$\endgroup$
  • $\begingroup$ Experimentally it goes down close to zero for large $x$. $\endgroup$ – joro Sep 8 '15 at 6:25
  • 1
    $\begingroup$ For $x$ near 1, good approximations of $g(x)$ and $g'(x)$ can be obtained using Euler-Maclaurin summation. For example, $$g(x) = -\tfrac12\ln2(x-1)^{-1} + \tfrac14\ln2 - (\tfrac1{48}+\tfrac1{24}\ln2)(x-1) + O((x-1)^2). $$ And similarly for $g'(x)$. It could be made into rigorous bounds. A bit tedious though. $\endgroup$ – Brendan McKay Sep 8 '15 at 7:28
  • 4
    $\begingroup$ It is equivalent (by easy manipulation) to say that the function $h(x)=(x-x^{-1})g(x)$ is increasing for $x>1$. I am not sure if that helps. $\endgroup$ – Neil Strickland Sep 8 '15 at 13:35
  • $\begingroup$ Another restatement can be done via the logarithmic derivative (en.wikipedia.org/wiki/Logarithmic_derivative). Your inequality is equivalent to $$ \mathrm{dlog} \bigl ( (x^3-x)^3 g(x) \bigr) > -\frac{4}{3}\frac{1}{x^3-x} $$ $\endgroup$ – Vít Tuček Sep 8 '15 at 13:56
13
$\begingroup$

I am not sure at all, please double check.

Denote $u(t)=(1+t)^{-1}$, it is a decreasing convex function on $[0,1]$, but $u(e^s)$ is concave on $(-\infty,0]$. As Neil Strickland notes, what we have to prove is that $(x-1/x)g(x)$ increases, or, if we denote $y=1/x$, that $(1/y-y)g(1/y)$ decreases. We have $$(1/y-y)g(1/y)=\sum_{k=0}^{\infty} \frac{y^{2k}-y^{2k+2}}{1+y^{2k+1}},$$ this is a Riemann sum of the function $u(t)$ corresponding to nodes $t_k=y^{2k}$, $t_0>t_1>\dots$, and intermediate points $s_k=y^{2k+1}=\sqrt{t_kt_{k+1}}\in [t_k,t_{k+1}]$. The Riemann sums converge to the integral, and they are always more than integral since $\int_a^b u(t)dt\leq (b-a)(u(a)+u(b))/2\leq (b-a)u(\sqrt{ab})$ as $u$ is convex and $u(e^s)$ is concave on $(-\infty,0]$. Next, we see that when $y$ increases, all nodes become closer to 1. It suggests to move nodes and see how the Riemann sum $R(t_0,t_1,\dots):=\sum (t_i-t_{i+1})u(\sqrt{t_it_{i+1}})$ behaves. Consider three consecutive nodes $a^2<b^2<c^2$ and increase $b$. What happens to $(b^2-a^2)u(ab)+(c^2-b^2)u(bc)$? Its derivative in $b$ equals $$ -\frac{(c-a)((a-c)^2+3(ac-b^2)+2(abc-b^3)(a+c)+ab^2c(ac-b^2))} {(ab+1)^2 (bc+1)^2}. $$ This is strictly negative if $b^2=ac$ (that holds in our case). It follows that $R(1,y^2,y^4,\dots)$ decreases when $y$ increases, as desired, by $$ \frac{d}{dy} R(1,y^2,y^4,\dots)=\sum_{k=1}^{\infty} 2ky^{2k-1}\frac\partial{\partial t_k} R(1,y^2,\dots)\geq 0. $$

$\endgroup$
  • 2
    $\begingroup$ It's worth to mention that by telescoping, we have $$\sum_{k\geq 0} \frac{y^{2k}-y^{2k+2}}{1+y^{2k+1}} = \frac{1}{1+y} + \sum_{k\geq 0} \frac{y^{2k+2}}{1+y^{2k+3}} - \frac{y^{2k+2}}{1+y^{2k+1}}$$ $$ = \frac{1}{1+y} + (1-y^2) \sum_{k\geq 0} \frac{y^{4k+3}}{(1+y^{2k+1})(1+y^{2k+3})}.$$ $\endgroup$ – Max Alekseyev Sep 8 '15 at 23:03
  • 1
    $\begingroup$ @Fedor, why $\int_{a}^{b}u(t)\leq (b-a)u(\frac{a+b}{2})$? This is not true in general for convex functions. this is also not true for $u(x)=(1+x)^{-1}$. Or am I missing something else? $\endgroup$ – Paata Ivanishvili Sep 9 '15 at 1:56
  • $\begingroup$ anyway, in your argument you do not need to use this opposite Hermite--Hadamard inequality :) $\endgroup$ – Paata Ivanishvili Sep 9 '15 at 3:55
  • 1
    $\begingroup$ Extremely clever! Seems like this trick with presenting the sum as an approximation of an integral is an instance of some powerful general method, is there such a thing? $\endgroup$ – მამუკა ჯიბლაძე Sep 9 '15 at 5:40
  • 2
    $\begingroup$ I consider generating functions as Riemann sums along geometric progressions every day. For example, this allows to get maybe the shortest proof of partition asymptotics $\log p(n)\leq 2\sqrt{\zeta(2) n}$. $\endgroup$ – Fedor Petrov Sep 9 '15 at 5:55
10
$\begingroup$

The inequality is equivalent to $$S := (x^2+1)g(x) + x(x^2-1)g'(x) > 0.$$ The left hand side here can be expanded to $$S = \sum_{k\geq 0} \frac{(x^2+1)(x^{2k+1}+1) - (2k+1)x^{2k+1}(x^2-1)}{(x^{2k+1}+1)^2} $$ $$= \sum_{k\geq 0} \frac{(x^2+1) - (2k+1)(x^2-1)}{x^{2k+1}+1} + \sum_{k\geq 0}\frac{(2k+1)(x^2-1)}{(x^{2k+1}+1)^2}.$$

Now, the first sum here simplifies to $$\sum_{k\geq 0} \frac{(x^2+1) - (2k+1)(x^2-1)}{x^{2k+1}+1} = \sum_{k\geq 0} \frac{(2k+2)-2k x^2}{x^{2k+1}+1}$$ $$=\sum_{k\geq 1} \left( \frac{2k}{x^{2k-1}+1} - \frac{2k x^2}{x^{2k+1}+1}\right)=(1-x^2)\sum_{k\geq 1} \frac{2k}{(x^{2k-1}+1)(x^{2k+1}+1)}.$$ Hence $$\frac{S}{x^2-1} = \sum_{k\geq 0}\frac{2k+1}{(x^{2k+1}+1)^2} - \sum_{k\geq 1} \frac{2k}{(x^{2k-1}+1)(x^{2k+1}+1)}$$ $$\geq \sum_{k\geq 0}\frac{2k+1}{(x^{2k+1}+1)^2} - \sum_{k\geq 1} \frac{2k}{(x^{2k}+1)^2} = \sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2}.$$ Here we used AM-GM inequality $x^{2k-1}+x^{2k+1}\geq 2x^{2k}$ and thus $$(x^{2k-1}+1)(x^{2k+1}+1)=x^{4k}+x^{2k+1}+x^{2k-1}+1\geq (x^{2k}+1)^2.$$ So it remains to prove that for $x>1$, $$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2} > 0.\qquad(\star)$$

UPDATE #1. Substituting $x=e^{2t}$, we have $$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2} = \sum_{k\geq 1} \frac{(-1)^{k-1}ke^{-2tk}}{4 \cosh(tk)^2} = \frac{1}{4}\sum_{k\geq 1} (-1)^{k-1}ke^{-2tk}(1-\tanh(tk)^2)$$ $$ = \frac{e^{2t}}{4(e^{2t}+1)^2} - \frac{1}{4}\sum_{k\geq 1} (-1)^{k-1}ke^{-2tk} \tanh(tk)^2.$$

UPDATE #2. The proof of $(\star)$ is given by Iosif Pinelis.

$\endgroup$
  • 1
    $\begingroup$ That last sum seems to rapidly approach $1/16$ as $x \to 1+$, e.g. in GP $$ $$ plot(x=1.001,2,suminf(k=1,(-1)^(k-1)*k/(x^k+1)^2) $\endgroup$ – Noam D. Elkies Sep 8 '15 at 20:10
  • $\begingroup$ @NoamD.Elkies: The limit at $x\to 1^+$ can be easily seen from the UPDATE#1. Although I've not yet completed the proof of the original statement, it seems to be close. $\endgroup$ – Max Alekseyev Sep 8 '15 at 22:46
  • $\begingroup$ I do accept your answer (of course in addition with final steps made by Iosif) $\endgroup$ – Paata Ivanishvili Sep 9 '15 at 3:53
9
$\begingroup$

Writing $k=\sum_{j=1}^k 1$ and then summing by parts, one has $$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2} =\sum_{j\ge 1}(-1)^{j-1}s_j =\sum_{k\ge0}[s_{2k+1}-s_{2k+2}], $$ where $$s_j:=\sum_{k=j}^\infty(-1)^{k-j} a(k)=\sum_{m\ge0}(-1)^m a(m+j) =\sum_{k\ge0}[a(2k+j)-a(2k+1+j)] $$ and $a(q):=\dfrac1{(x^q+1)^2}$. So, $$s_j-s_{j+1}=\sum_{k\ge0}[a(2k+j)-2a(2k+1+j)+a(2k+2+j)]>0, $$ since $a''>0$ and hence $a$ is strictly convex. So, $\sum_{k\geq 1} \dfrac{(-1)^{k-1}k}{(x^{k}+1)^2}>0$, and the result follows by Max Alekseyev's answer.

$\endgroup$
  • 1
    $\begingroup$ Great! By the same argument, $\sum_{k\geq 1} (-1)^{k-1} k f(k)$ is positive for any positive convex function $f(k)$. $\endgroup$ – Max Alekseyev Sep 9 '15 at 1:50
  • 1
    $\begingroup$ Right. Also, I think that iterating this argument one can show that, if $f$ has a higher-order convexity property, then instead of the factor $k$ one can use a polynomial in $k$ of the corresponding degree. $\endgroup$ – Iosif Pinelis Sep 9 '15 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.