To check convergence of series $\sum_{1}^{\infty}\frac{\ln(n)}{n(n+1)}$

You can do a bound test for large $n$.

For $n>N$ where $N$ is some bound, the following is true: $$\frac{\ln n}{n(n+1)}<n^{-2+\epsilon}$$ where $1>\epsilon>0$.

We know that $\sum n^k$ converges for $k<-1$. I used the fact that $\mathcal{O}(\ln n)<\mathcal{O}(n^\epsilon)$ for any $\epsilon>0$.

Here's another way to prove it. We know that $\ln n <\sqrt{n}$ starting from some $n$.

Since $\frac{\ln n}{n(n+1)}<\frac{\sqrt{n}}{n(n+1)}=\frac1{\sqrt{n}(n+1)}\sim \frac1{n^{3/2}}$ and we know that series $\sum_{k=1}^\infty \frac1{k^\alpha}$ converges if $\alpha>1$. Therefore $$\sum_{n=1}^\infty \frac{\ln n}{n(n+1)}$$ converges.