Does $\sum\limits_n \log\left(1+{1\over n}\right)$ diverge or converge? How do I find out if $\sum\limits_n\log(1+{1\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.
$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n}=\log(N+1).$$

If you want to use a comparison, notice that since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have:
$$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1/2}$$
by Jensen's inequality, hence:
$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)\geq 2\sum_{n=1}^{N}\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$
Another way to think of this problem is that $\sum log(1+1/n)$ = $\log \prod(1+1/n)$ = $\log \prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.