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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

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  • $\begingroup$ Experimentally it goes down close to zero for large $x$. $\endgroup$
    – joro
    Commented Sep 8, 2015 at 6:25
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    $\begingroup$ 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
    Commented Sep 8, 2015 at 7:28
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    $\begingroup$ 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
    Commented Sep 8, 2015 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
    Commented Sep 8, 2015 at 13:56

4 Answers 4

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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 $$ \frac{d}{dy} R(1,y^2,y^4,\dots)=\sum_{k=1}^{\infty} 2ky^{2k-1}\frac\partial{\partial t_k} R(1,y^2,\dots)\geq 0. $$

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    $\begingroup$ 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
    Commented Sep 8, 2015 at 23:03
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    $\begingroup$ @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
    Commented Sep 9, 2015 at 1:56
  • $\begingroup$ anyway, in your argument you do not need to use this opposite Hermite--Hadamard inequality :) $\endgroup$
    – Paata Ivanishvili
    Commented Sep 9, 2015 at 3:55
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    $\begingroup$ 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
    Commented Sep 9, 2015 at 5:40
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    $\begingroup$ 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
    Commented Sep 9, 2015 at 5:55
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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.

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    $\begingroup$ 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
    Commented Sep 8, 2015 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
    Commented Sep 8, 2015 at 22:46
  • $\begingroup$ I do accept your answer (of course in addition with final steps made by Iosif) $\endgroup$
    – Paata Ivanishvili
    Commented Sep 9, 2015 at 3:53
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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.

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    $\begingroup$ 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
    Commented Sep 9, 2015 at 1:50
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    $\begingroup$ 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
    Commented Sep 9, 2015 at 2:01
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Now I found \begin{align*}\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)}\\ &=\sum_{k=1}^{+\infty}k\left(\dfrac{1}{x^{2k-1}+1}-\dfrac{1}{x^{2k+1}+1}\right)^2\\ &\ge 0 \end{align*}

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  • $\begingroup$ Nice observation! The idea of using two consecutive terms of the sum is amazing which, I believe, it first appeared in Fedor Petrov’s answer. $\endgroup$
    – Paata Ivanishvili
    Commented Jan 4 at 14:31

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