Problem :
Let $a>M$ be positives integers then :
$$\sum_{i=1}^{n}\left(\frac{a}{i\ln\left(i\right)+1}\right)!-n> a!$$
I know the formula $(n+1)!=(n+1)n!$ but it seems an additive analogue to this setting $a=n+1$.
We have also $a,M$ as above :
$$\sum_{i=2}^{n}\left(\frac{a}{\ln^{2}\left(n\right)\cdot i\ln\left(i\right)+1}\right)!-n+\ln\left(n\right)>0$$
My attempt :
For the second problem :
We have for $x>0$ then :
$$x!\geq f\left(x\right)=1+\frac{x\left(x-1\right)}{cx+1}\tag{I}$$
Where : $c=\frac{\gamma}{1-\gamma}$ and $\gamma$ is the Euler-Mascheroni constant .
Using $I$ we need to show :
$$\left(\frac{a}{\ln^{2}\left(n\right)\cdot2\ln\left(2\right)+1}\right)!+\sum_{i=3}^{n}\left(f\left(\frac{a}{\ln^{2}\left(n\right)\cdot i\ln\left(i\right)+1}\right)\right)-n+\ln\left(n\right)> 0$$
As $f(x)$ is convex on the domain we use Jensen's inequality :
We need to show :
$$\left(\frac{a}{\ln^{2}\left(n\right)\cdot2\ln\left(2\right)+1}\right)!+\left(n-3\right)f\left(\frac{\left(\displaystyle \sum_{i=3}^{n}\frac{a}{\ln^{2}\left(n\right)\cdot i\ln\left(i\right)+1}\right)}{n-3}\right)-n+\ln n>0$$
Which is true for large $a,n$
I'm stuck here ...
How to show it ?