How to determine whether $\sum_{n=1}^{\infty}\ln\left(\frac{n+2}{n+1}\right)$ converges or diverges. I am trying to find whether $\sum_{n=1}^{\infty}\ln\left(\frac{n+2}{n+1}\right)$ converges or diverges. I used the limit test, and it comes out as inconclusive since $\lim_{n\rightarrow\infty}\ln\left(\frac{n+2}{n+1}\right) = 0$. When I put it into wolfram, it states the series diverges by comparison test. But I don't know how to set up the comparison test (what series to compare it to). All help in solving this would be greatly appreciated, thanks.
 A: Hints:
$\ln\frac{n+2}{n+1}=\ln(n+2)-\ln(n+1)$.
Then $$\sum_{n=1}^{\infty}\ln\frac{n+2}{n+1}=\lim_{m\to \infty}\sum_{n=1}^{m}\ln\frac{n+2}{n+1}=\lim_{m\to \infty} [\ln(m+2)-\ln2]=+\infty$$
A: Observe that: $\displaystyle \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n+1}\right)}{\dfrac{1}{n+1}} = 1$, and the harmonic series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n+1}$ diverges, so the series $\displaystyle \sum_{n=1}^\infty \ln\left(1+\dfrac{1}{n+1}\right)$ also diverges.
A: Since $\log \frac {n + 2} {n + 1} = \log (1 + \frac {1} {n + 1}) \sim \frac {1} {n +1}$, we have
$$\sum_{n = 1}^{\infty} \log \frac {n + 2} {n + 1} \sim \sum_{n = 1}^{\infty} \frac {1} {n +1}$$
which diverges.
A: A cool way think about it is this-
$\sum_{n=1}^{\infty}\ln\left(\frac{n+2}{n+1}\right)=\ln\left(\Pi_{n=1}^{\infty}\left(\frac{n+2}{n+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\Pi_{n=1}^{N}\left(\frac{n+2}{n+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\left(\frac 3 2\cdot\frac 4 3\cdot\dotso\cdot\frac{N+1}N\cdot\frac{N+2}{N+1}\right)\right)=\ln\left(\lim_{N\rightarrow\infty}\frac{N+2}2\right)=\infty$

In general, when you have an infinite sum of $\ln$s, it is sometimes beneficial to look at it as $\ln$ of the product.
$\sum_{n=1}^{\infty}\ln\left(a_n\right)=\ln\left(\Pi_{n=1}^{\infty}\left(a_n\right)\right)$
A: We know that $\ln(1+x)\leq x$ for all $x>-1$ which gives $-\ln(1+x)\geq -x$. Therefore we have that, for all $n\in\mathbb{N}$,
$$\ln\left(\frac{n+2}{n+1}\right)=-\ln\left(\frac{n+2-1}{n+2}\right)$$
$$\implies\ln\left(\frac{n+2}{n+1}\right)=-\ln\left(1-\frac{1}{n+2}\right)\geq \frac{1}{n+2}$$
But the series $\sum_{n=1}^\infty \frac{1}{n+2}$ diverges.
