How do you show $ \lim_{n\to \infty} {\frac{\ln(1)+\ln(2)+\ldots+\ln(n)}{n}} = \infty$? $$ \lim_{n\to \infty} {\frac{\ln(1)+\ln(2)+\ldots+\ln(n)}{n}} = \infty$$
I know I should show that it is greater then something that approaches to $\infty$ but I don't see what.
 A: If we set $a_n=n!$, we have $\frac{a_{n+1}}{a_n}=(n+1)$, so:
$$ \lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=+\infty $$
implies:
$$ \lim_{n\to +\infty} \sqrt[n]{n!} = +\infty $$
and by switching to logarithms we prove our claim. As an alternative approach:
$$\frac{1}{n}\sum_{k=1}^{n}\log k = \log n+\frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}$$
but the last sum is a Riemann sum, and:
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}=\int_{0}^{1}\log(x)\,dx = -1.$$
A: The sum in the numerator is equal to $\ln(n!)$. For a very rough estimate we have $n!\geq n^{n/2}$ (you can prove this by induction), so $\ln(n!)\geq \frac{n}{2}\ln(n)$. This gets you what you want.
A: $$  \log(1) + ... + \log(n) \geq \int_1^n \log(x) ~ dx $$
A: Hint:
It will suffice to show that
$$\lim_{n \to {\infty}} {{\ln(n!)} \over n}=\infty$$
To prove this, use
$${\ln(n!) \over n}=\ln({(n!)}^{1/n})$$
A: Throw away the first $\sqrt{n}$ terms in the numerator (they are all nonnegative, so this is a lower bound for the sum).  The remaining terms are all at least $\log \sqrt{n} = \tfrac12 \log n$, and there are at least $n - \sqrt{n}$ of them, so the numerator is at least $\tfrac12 (n-\sqrt{n})\log n$.  Can you see why the ratio goes to infinity?
A: Using Stirling's Formula,
\begin{align}
\lim_{n\to\infty}\frac1n\ln{(n!)}&=\lim_{n\to\infty}\frac1n\ln{(\sqrt{2\pi n}(\frac n e)^n )}\\&=\ln[\lim_{n\to\infty}(2\pi n)^{\frac1{2n}}\cdot\frac{n}e]\\&=\ln[\lim_{n\to\infty}(2n)^{\frac{1}{2n}}\cdot\lim_{n\to\infty}\pi^{\frac{1}{2n}}\cdot\lim_{n\to\infty}\frac n e]\\&=\lim_{n\to\infty}\ln(1\cdot1\cdot n)-1\\&=\infty
\end{align}
A: Suppose $n$ is even. Then in the numerator, at lease $n/2$ of the terms are $>\ln (n/2).$ Thus
$$\frac{\ln 1 + \cdots + \ln n}{n} > \frac{(n/2)\ln (n/2)}{n} = (1/2)\ln (n/2) \to \infty$$
as $n\to \infty$ through even $n.$ This gives the idea; handling odd $n$ is a small step from the above.
