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) 



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> My question is -> how authors deduced that $c\_k,j \leq (2D)^{3Dn} ( n! / (k)^{ n+1} )^{s+1-3D}$ , for n large. 
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[!inequalities in which I have question](https://i.stack.imgur.com/u0IZR.jpg) ][1](https://i.stack.imgur.com/u0IZR.jpg)


[!defination of $c\_{k,j}$](https://i.stack.imgur.com/eo863.jpg)][2](https://i.stack.imgur.com/eo863.jpg)


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**](http://dx.doi.org/10.1112/S0010437X1900722X), Compos. Math. 155, No. 5, 938-952 (2019). [ZBL1430.11097](https://zbmath.org/?q=an:1430.11097).



Unable to deduce an inequality in paper on odd zeta values of Fischler, Sprang and Zudilin
**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.


