I am a masters student and I am interested in number theory. Due to lockdown I have a lot of time and I thought of reading a research paper in Number theory which is " Many Odd zeta values are irrational by " Stephane Fischler, Johannes Sprang and Wadim Zudilin.

I have a question on page 8 just after the (3.6)

My question is -> how authors deduced that $c_k,j \leq (2D)^{3Dn} ( n! / (k)^{ n+1} )^{s+1-3D}$ , for n large.

!inequalities in which I have question ]1

!defination of $c_{k,j}$]2

I am not able to understand how authors came to this conclusion. I think probably using $c_k, j$ 's definition one could get it. I can divide and multiply by $(2)^{3Dn} $ and use that s+1> 3D and that there are n+1 terms in denominator raised to exponent s+1 . These things indicate me that definition of $c_{j, n} $ would be used but I am not getting exact given inequality.

Edit -> Unfortunately there is one more question I am having. I am not able to derive the inequality which is just after the line Using (3.1) and Stirling Approximation .

Can someone please tell how it will be derived.

I have tried it many times.

Fischler, Stéphane; Sprang, Johannes; Zudilin, Wadim, Many odd zeta values are irrational, Compos. Math. 155, No. 5, 938-952 (2019). ZBL1430.11097.

  • $\begingroup$ I inserted the citation for the paper. Remark: Zudilin sometimes posts here! $\endgroup$ – Gerald Edgar Apr 21 at 15:04
  • $\begingroup$ @GeraldEdgar Thanks!! $\endgroup$ – Tim Green Apr 21 at 17:03

1. Let us prove the first inequality. From the definition of $c_{k,j}$, it is clear that $$c_{k,j}\leq D^{3Dn} n!^{s+1-3D} (k+3n+1)^{3Dn+1}k^{-(s+1)(n+1)},$$ hence it suffices to verify that $$(k+3n+1)^{3Dn+1}\leq 2^{3Dn}k^{3D(n+1)}.$$ As explained in the paper, $k$ is much larger than $n$, and $n$ is itself large. Hence $k+3n+1$ is at most $2k$, and it suffices to show that $$(2k)^{3Dn+1}\leq 2^{3Dn}k^{3D(n+1)}.$$ This reduces to $2\leq k^{3D-1}$, which is obvious.

2. Let us prove the second inequality. We shall abbreviate $A(\varepsilon)-\varepsilon$ by $B(\varepsilon)$. By Stirling's approximation, $n!<(n/e)^{n+1}$ for $n$ large, hence it suffices to show that $$2(2D)^{3Dn}\frac{(n/e)^{(n+1)(s+1-3D)}}{(B(\varepsilon)n)^{(n+1)(s+1-3D)-2}}\leq\left(\frac{2D}{eB(\varepsilon)}\right)^{sn/2}.$$ Equivalently, $$2(2D)^{3Dn}\frac{(B(\varepsilon)n)^2}{(eB(\varepsilon))^{(n+1)(s+1-3D)}}\leq\left(\frac{2D}{eB(\varepsilon)}\right)^{sn/2}.$$ For this it suffices that $$2(2D)^{3Dn}\leq(2D)^{sn/2}\tag{1}$$ and $$(eB(\varepsilon))^{sn/2}(B(\varepsilon)n)^2\leq(eB(\varepsilon))^{(n+1)(s+1-3D)}.\tag{2}$$ Both $(1)$ and $(2)$ follow from the fact that $s>6D$ and $n$ is large.

| cite | improve this answer | |
  • $\begingroup$ I have actually 2 questions here. Probably, you didn't saw the Edit . Can you please answer that too $\endgroup$ – Tim Green May 4 at 18:51
  • $\begingroup$ I have completely understood your this answer but can you please prove that inequality too. It was asked earlier on same day the question was asked but unfortunately you didn't saw it. $\endgroup$ – Tim Green May 4 at 19:14
  • $\begingroup$ @YannicMuller: Please restrict to one question per post. This is site policy. $\endgroup$ – GH from MO May 4 at 19:28
  • $\begingroup$ I will keep that in mind but can you please answer this question. I have spent enough time trying to prove that inequality and also if I ask a new question now it will have a lot of common with this question. Can you please answer only for this time? $\endgroup$ – Tim Green May 4 at 19:33
  • 1
    $\begingroup$ @YannicMuller: I added a proof of the second inequality. $\endgroup$ – GH from MO May 4 at 20:03

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.