Question: Hello I have the following formula that I need to transform to a normal form required for the resolution rule. I understand that this from is a sum of OR's as in: 
$(V1 \lor V2 ) \land ( V3 \lor V4 ) )$

The formula:

$\forall x \forall y \bigl[P(x,y) \Rightarrow \exists z [Q(y,z] \land \lnot R(x,z)]\bigl] \lor \forall z Q(x,y,z)$

so, I use Skolem to substitute $\exists z$ with a constant A, then removed the universal quantifiers and following the general logic rules I came up with this:

$\bigl( \lnot P(x,y) \lor Q(y,A) \lor Q(x,y,z) \bigl) \land \bigl( \lnot P(x,y) \lor R(x,A) \lor Q(x,y,z) \bigl)$

Or should I first move the right most $\forall z$ to the left and then through skolem substitute the $\exists z$ with a function f(z) ?

$\bigl( \lnot P(x,y) \lor Q(y,f(z)) \lor Q(x,y,f(z)) \bigl) \land \bigl( \lnot P(x,y) \lor R(x,f(z)) \lor Q(x,y,f(z)) \bigl)$

Which solution is correct?

Answer: 
By Stolz-Cesaro

$$\frac{a\_{n}}{b\_{n}} = \lim\_{n\rightarrow \infty} \frac{\sum^{n}\_{k=1}\frac{a\_{k}}{k}}{\ln (n)}$$

then

$$\lim\_{n\rightarrow \infty}\frac{a\_{n+1}-a\_{n}}{b\_{n+1}-b\_{n}} =\lim\_{n\rightarrow \infty}\frac{\frac{a\_{n+1}}{n+1}}{\ln (n+1)-\ln n}=\lim\_{n\rightarrow \infty}\frac{\frac{n}{n+1}a\_{n+1}}{\ln \left(1+\frac1n\right)^n} \to a$$
What the author of the solution is using is that $\lim\_{n \to \infty}a\_{n+1} = \lim\_{n \to \infty}a\_n$.

Therefore, we may rewrite the last step in a more rigorous way as 

$\lim\_{n \to \infty}\frac{a\_{n+1}}{(n+1)ln(1+\frac{1}{n})} = \frac{\lim\_{n \to \infty}a\_{n+1}}{lim\_{n \to \infty}ln(1+\frac{1}{n})^{n+1}} = \frac{a}{\log e} = a$, where I also used the fact that $ lim\_{n \to \infty}(1+\frac{1}{n})^{n+1} = e$

It doesn't have to be the case that $a\_{n+1} = a\_n$ for the solution to work.