Determine $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$ and $\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$ I want to use Stirlings Formula $\lim_{n\to \infty} {n!} \sim \lim_{n \to \infty} \frac{n^n}{e^n}\cdot \sqrt{n}$ to evaluate the following limits:
$$\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}}$$
$$\lim_{n\to \infty}[ \frac{n+\sum_{k=1}^{n-1}\log(k)}{\log(n)} +1 -n]$$
For the first one I feel like some algebraic manipulation would yield $\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} = \frac{e}{n^{^1/n}}$ which would imply $\lim_{n\to \infty} \frac{n}{\sqrt[n]{n!}} = e$ but can I just do equivalence transformation with an asymptotical equality, as opposed to a precise equality?
The second one looks like Taylor Series development, but I don't see the path...

Stirling is for whimps. Just kidding, but let's have a go without it. In the second problem the expression equals
$$\tag 1 \frac{n+(\sum_{k=1}^{n-1} \ln k) +(1-n) \ln n}{\ln n}.$$
As the denominator $\to \infty,$ we're entitled to see what Stolz-Cesaro has to say. So consider the quotient of differences:
$$\frac{1 + \ln n + (1-(n+1)) \ln (n+1) -(1-n) \ln n }{\ln (n+1) - \ln n}.$$
Now $\ln (n+1)= \ln n + \ln(1+1/n).$ Use that in the above to get
$$\frac{1-n\ln (1+1/n)}{\ln (1+1/n)}.$$
Because $\ln (1+1/n)= 1/n-1/(2n^2) + O(1/n^3),$ the above equals
$$\frac{1/(2n) + O(1/n^2)}{1/n-1/(2n^2) + O(1/n^3)} = \frac{1/2 +O(1/n)}{1-1/(2n) +O(1/n^2)} \to \frac{1}{2}.$$
By Stolz-Cesaro, the limit of $(1)$ is $1/2.$
