Calculate $\lim_{n \to \infty} \ln \frac{n!^{\frac{1}{n}}}{n}$ How can I calculate the following limit? I was thinking of applying Cesaro's theorem, but I'm getting nowhere. What should I do?
$$\lim_{n \to \infty} \ln \frac{n!^{\frac{1}{n}}}{n}$$
 A: First let's write $$\ln \frac{n!^{\frac{1}{n}}}{n} = \frac{1}{n} (\ln n! - n\ln n) = \frac{a_n}{b_n}$$ where $a_n = \ln n! - n\ln n$ and $b_n = n$. 
Then 
$$\begin{align*} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} & = \quad  \frac{\ln (n+1)! - (n+1) \ln(n+1) - (\ln n! - n\ln n)}{1} \\  & =  \quad \ln(n+1) - (n+1)\ln(n+1) + n \ln n \\ & =  \quad -\ln(1 + 1/n)^n \\ & \longrightarrow  -1 \ \text{ as } n \to \infty \end{align*}$$
Hence by the Cesàro theorem (a.k.a. Stolz-Cesàro theorem) 
$$ \lim_{n\to\infty} \ln \frac{n!^{\frac{1}{n}}}{n} = \lim_{n\to\infty} \frac{a_n}{b_n}= \lim_{n\to\infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}  =  -1$$
A: $$\ln(n!^{\frac1{n}}) = \frac1{n} \ln(n!) = \frac1{n} \sum_{k = 1}^n \ln(k) = \frac{\sum_{k = 1}^n \ln(k)}{n}$$ 
Then: 
$$\ln \frac{n!^{\frac1{n}}}{n} = \ln(n!^{\frac1{n}}) - \ln(n) =  \frac{\sum_{k = 1}^n \ln(k)}{n} - \ln(n) = \frac{\sum_{k = 1}^n \ln(k) - n \ln(n)}{n} = \frac{\sum_{k = 1}^n \left( \ln(k) - \ln(n) \right)}{n} = \frac{\sum_{k = 1}^n \ln(\frac{k}{n})}{n}$$
Note that: $$\lim_ n \frac{1}{n} \sum_{k = 1}^n \ln(\frac{k}{n}) = \int_0^1 \ln x dx = -1 - \lim_{x \to 0} (x \ln x - x) = -1$$
Therefore:
$$\lim_n \ln \frac{n!^{\frac1{n}}}{n} = -1$$
A: Here are some transformations:\begin{eqnarray*}
\ln \left( \frac{n!^{1/n}}{n}\right)  &=&\frac{1}{n}\ln n!-\ln n=\frac{1}{n}%
\sum_{k=1}^{n}\ln k-\ln n=\frac{1}{n}\sum_{k=1}^{n}\ln \left( n\frac{k}{n}%
\right) -\ln n \\
&=&\frac{1}{n}\sum_{k=1}^{n}(\ln n+\ln \left( \frac{k}{n}\right) )-\ln n \\
&=&\frac{1}{n}\sum_{k=1}^{n}\ln n+\frac{1}{n}\sum_{k=1}^{n}\ln \left( \frac{k%
}{n}\right) -\ln n \\
&=&\frac{1}{n}n\ln n+\frac{1}{n}\sum_{k=1}^{n}\ln \left( \frac{k}{n}\right)
-\ln n \\
&=&\frac{1}{n}\sum_{k=1}^{n}\ln \left( \frac{k}{n}\right) .
\end{eqnarray*}
Can you take it from here?
A: We have:
$$\log\frac{n!^{\frac{1}{n}}}{n}=-\log n+\frac{1}{n}\sum_{k=1}^{n}\log k, \tag{1}$$
and summation by parts gives:
$$\sum_{k=1}^{n}\log k = n \log n-\sum_{k=1}^{n-1} k \log\left(1+\frac{1}{k}\right)\tag{2}$$
hence we just have to compute:
$$\lim_{n\to +\infty}-\frac{1}{n}\sum_{k=1}^{n-1}k\int_{k}^{k+1}\frac{dx}{x}\tag{3}$$
and we may exploit the Hermite-Hadamard inequality, since $\frac{1}{x}$ is a convex function on $\mathbb{R}^+$.
That gives:
$$\sum_{k=1}^{n-1}k\int_{k}^{k+1}\frac{dx}{x}\leq \frac{1}{2}\sum_{k=1}^{n-1}\left(1+\frac{k}{k+1}\right)=n+O(\log n)\tag{4} $$
that together with the trivial:
$$ \sum_{k=1}^{n-1}k\int_{k}^{k+1}\frac{dx}{x}\geq \sum_{k=1}^{n-1}\frac{k}{k+1}=n-O(\log n)\tag{5} $$
gives that the value of our limit is just $\color{red}{-1}$.
A: Another approach using Stirling's approximation: 
\begin{align*}
\lim_{n \to \infty} \ln \frac{n!^{\frac{1}{n}}}{n} &\approx \lim_{n \to \infty} \ln{ \frac{ \sqrt[n]{\sqrt{2\pi n}(\frac{n}{e})^n}}{n}}\\
&= \lim_{n \to \infty} \ln{ \sqrt[2n]{2\pi n}(\frac{n}{e}) } - \ln n\\
&= \lim_{n \to \infty} \frac{\ln 2\pi n}{2n} + \ln n - \ln e - \ln n\\
&\overset{(1)}{=} 0 - 1 = -1
\end{align*}
$(1)$ uses the fact that $\lim_{n \to \infty} \frac{\ln n} {n} = 0$
