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...
 A: If you consider that
$$ \int_{0}^{1}\log(x)\,dx = \left[x\log(x)-x\right]_{0}^{1} = -1 \tag{1}$$
by the definition of the Riemann integral in terms of Riemann sums you also get that
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{k}{n}\right) = -1 \tag{2} $$
hence it follows that
$$ \log(n!)-n\log(n)+n = O(\log n)\tag{3} $$
and this identity proves that $\lim_{n\to +\infty}\frac{n}{\sqrt[n]{n!}}=\color{red}{e}$. In order to solve your second limit you need to find the hidden constant in the RHS of $(3)$. For such a purpose, one may estimate the integral $\int_{0}^{1}\log(x)\,dx$ by using trapezoids and carefully bounding the error term. That leads to:
$$ \log(n!) = \left(n+\frac{1}{2}\right)\log n-n+O(1) \tag{4} $$
settling both your limits and a rudimentary form of Stirling's inequality up to the constant $\sqrt{2\pi}$.
A: If you are unsure about what you are allowed to do with equivalents, then go back to the equivalent characterization as first-order Taylor expansion (in particular, when you write Stirling's formula, you should not have $\lim$ on either side: this is mixing two different notions; you are also missing a constant):
$$
n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\frac{n^n}{e^n} \quad\Longleftrightarrow\quad n! = \sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right) \tag{1}
$$
With (1), we can then write
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{n}{\sqrt[n]{\sqrt{2\pi n}\frac{n^n}{e^n} + o\left(\sqrt{n}\frac{n^n}{e^n}\right)}}= \frac{n}{\frac{n}{e}(2\pi n)^{1/2n}\sqrt[n]{1 + o\left(1\right)}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
$$
Now, we can observe that
$$
(2\pi n)^{1/2n} = \exp\left(\frac{1}{2n}\ln(2\pi n)\right) \xrightarrow[n\to\infty]{} e^0 = 1
$$
and
$$
(1+o(1))^{1/n} = \exp\left(\frac{1}{n} \ln(1+o(1))\right) \xrightarrow[n\to\infty]{} e^0 = 1$$
so that
$$
\frac{n}{\sqrt[n]{n!}}
= \frac{e}{(2\pi n)^{1/2n}}\cdot \frac{1}{(1+o(1))^{1/n}}
\xrightarrow[n\to\infty]{} \boxed{e}
$$

For the second: establishing that $$\sum_{k=1}^{n-1} \log k = \log \prod_{k=1}^{n-1} k = \log((n-1)!)$$ and applying Stirling's formula will do the trick:
$$
\log n!  = n\log n - n + \frac{1}{2}\log (2\pi n) + o(1) \tag{2}
$$
from which
$$\begin{align}
\frac{n+\log( (n-1)! )}{\log n} +1 - n
&= 
\frac{n+\log( n! ) - \log n}{\log n} +1 - n
= 
\frac{n+\log( n! )}{\log n} - n
\\
&= \frac{n\log n + \frac{1}{2}\log (2\pi n) + o(1)}{\log n} - n\\
&= \frac{\frac{1}{2}\log (2\pi n) + o(1)}{\log n} \xrightarrow[n\to\infty]{} \cdots
\end{align}$$
A: 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 $1/2.$
