No; it diverges. First of all, note that $\dfrac n{n(n+1)}=\dfrac1{n+1}$. On the other hand, the series $\sum_{n=0}^\infty\frac1{n+1}$ is the harmonic series, which diverges.
The series that you wrote after editing your question is $\displaystyle\sum_{n=1}^\infty\frac n{(2n-1)2n}=\sum_{n=1}^\infty\frac1{4n-2}$ which diverges too, since $(\forall n\in\mathbb N):\dfrac1{4n-2}\geqslant\dfrac14\times\dfrac1n$ and since the harmonic series diverges.
No; it diverges. First of all, note that $\dfrac n{n(n+1)}=\dfrac1{n+1}$. On the other hand, the series $\sum_{n=0}^\infty\frac1{n+1}$ is the harmonic series, which diverges.
The series that you wrote after editing your question is $\displaystyle\sum_{n=1}^\infty\frac n{(2n-1)2n}=\sum_{n=1}^\infty\frac1{4n-2}$ which diverges too, since $(\forall n\in\mathbb N):\dfrac1{4n-2}\geqslant\dfrac14\times\dfrac1n$ and since the harmonic series diverges.