Three logarithmic series

1. Series $\sum u_n$. For $n\ge1$,
$$u_n=\ln\!\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln n.$$
The partial sum from $n=1$ to $N$ telescopes:
$$S_N=\sum_{n=1}^{N}(\ln(n+1)-\ln n)=\ln(N+1)-\ln 1=\ln(N+1).$$
As $N\to+\infty$, $S_N\to+\infty$: the series diverges.

2. Series $\sum v_n$. The term $v_n$ is defined for $n\ge2$. For $n\ge2$,
$$v_n=\ln\!\left(\frac{n-1}{n}\right)=\ln(n-1)-\ln n.$$
The partial sum from $n=2$ to $N$ telescopes:
$$T_N=\sum_{n=2}^{N}(\ln(n-1)-\ln n)=\ln 1-\ln N=-\ln N.$$
As $N\to+\infty$, $T_N\to-\infty$: the series diverges (to $-\infty$).

3. Series $\sum w_n$. The term $w_n$ is defined for $n\ge2$. Split the partial sum into blocks:
$$w_{2p}=\ln\!\left(\frac{2p+1}{2p}\right),\qquad w_{2p+1}=\ln\!\left(\frac{2p}{2p+1}\right).$$
Note that $w_{2p}+w_{2p+1}=0$ for every $p\ge1$, since $\frac{2p+1}{2p}\cdot\frac{2p}{2p+1}=1$.

Odd partial sums. For $N\ge1$:
$$Z_{2N+1}=\sum_{n=2}^{2N+1}w_n=\sum_{p=1}^{N}(w_{2p}+w_{2p+1})=0.$$

Even partial sums. For $N\ge1$:
$$Z_{2N}=\sum_{n=2}^{2N}w_n=\ln\!\left(\frac{3}{2}\right)+\sum_{p=2}^{N}(w_{2p-1}+w_{2p}).$$
But for $p\ge2$, $w_{2p-1}+w_{2p}=\ln\!\left(\frac{2p-1}{2p}\right)+\ln\!\left(\frac{2p+1}{2p}\right)$ — the cancellation now leaves $\ln\frac{2N+1}{2N}$:
$$Z_{2N}=\ln\frac{3}{2}+\ln\frac{3}{4}+\ln\frac{5}{4}+\cdots+\ln\frac{2N+1}{2N}=\ln\frac{2N+1}{2N}.$$
Indeed the product telescopes:
$$\frac{3}{2}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{5}{6}\cdots\frac{2N+1}{2N}=\frac{2N+1}{2N}.$$
As $N\to+\infty$, $Z_{2N}=\ln\!\left(1+\frac{1}{2N}\right)\to0$.

Conclusion. Both $Z_{2N}$ and $Z_{2N+1}$ tend to $0$, so the sequence of partial sums $(Z_N)$ converges to $0$. Hence the series $\sum w_n$ converges with sum $0$.