TITLE: Prove divergence of sum $\ln(1+1/n)$
QUESTION [5 upvotes]: We have seen the harmonic series is a divergent series whose terms approach $0$. Show that $$\sum_{n = 1}^\infty \text{ln}\left(1 + \frac{1}{n}\right)$$ is another series with this property. 

Denote $a_n = \text{ln}\left(1 + \frac{1}{n}\right).$ Then, $$\lim_{n \to \infty}\text{ln}\left(1 + \frac{1}{n}\right) = \text{ln}\left(1 + \lim_{n \to \infty}\frac{1}{n}\right) = 0,$$ since $\text{ln}(x)$ is a continues function on its domain. However, I can't seem to prove that the series is divergent. I know the proof for the harmonic series (which is quite clever in my opinion), and I wondered whether this question required a similar approach?

REPLY [4 votes]: For all $n\in\mathbb{N}^\star$ :
$$\ln\left(1+\frac{1}{n}\right)=\ln\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln(n)$$
so that :
$$\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)-\ln(1)=\ln(N+1)\underset{N\to\infty}{\longrightarrow}+\infty$$
and the divergence of the series $\sum_{n\ge1}\ln\left(1+\frac{1}{n}\right)$ is proved.
Note that this gives us a proof (one of the easiest ones) of the divergence of the harmonic series, since :
$$\forall n\in\mathbb{N}^\star,\,\ln\left(1+\frac{1}{n}\right)\le\frac{1}{n}$$