I want to calculate $\lim_{n\to\infty}\frac{\log n!}{n\log n}$. I know this is a duplicate and I read its equal to $1$, however I can't seem to find the problem in the below calculation:
$$\lim_{n\to\infty}\frac{\log n!}{n\log n} = \lim_{n\to\infty} \frac{\log1 +\log2+\dots+\log n}{n\log n}$$
Now if we split the limits, we get:
$$\lim_{n\to\infty} \frac{\log1}{n\log n}+\dots+\lim_{n\to\infty} \frac{\log n}{n\log n}$$
Then each term will give $0$, and thus answer should be $0$.
HINT
This method is not valide since you are summing up infintely many terms.
As an alternative refer to Stolz-Cesaro
$$\lim_{n\to\infty}\frac{\log n!}{n\log n} = \lim_{n\to\infty}\frac{\log (n+1)!-\log n!}{(n+1)\log (n+1)-n\log n}$$