Why these two series are convergent or divergent? I do not understand why $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac1k$$ is divergent but the other series $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}k$$ is convergent. For both cases the $\displaystyle\lim_{n \to +\infty} z_{n} = 0$. Could you explain please?
I prefer to come up with an all-inclusive test of convergence that accepts the second but rejects the first. Thanks. 

OP wants a single test that can distinguish convergence and divergence of the 2 aforementioned series. So far all the answers used Cauchy's convergence test for first series and are essentially correct, albeit the presentations are questionable.
Actually Cauchy's test can also be applied on second series with some extra handling. Please refer to Wikipedia article for the variables used here. We need to prove for arbitrary positive real number $\epsilon$ there exists natural number $N$ so that  for any $n \gt N$ and $p \ge 1$, $$-\epsilon < \sum\limits_{n+1}^{n+p} \frac{(-1)^{k+1}}{k} < \epsilon$$
We pick $N$ so that $\frac{1}{N}$ is largest value smaller than $\epsilon$, i.e. $N=\lceil\frac{1}{\epsilon}\rceil$ (bracket means ceiling). Here 4 cases will need to be presented depending on evenness of $n$ and $p$.
1. $n$ even, $p$ even: (implying $n+1$ is odd and $n+p$ is even)
With different grouping on the terms,
$$\sum\nolimits = (\frac{1}{n+1}-\frac{1}{n+2}) + {...} + (\frac{1}{n+p-1}-\frac{1}{n+p}) > 0 > -\epsilon$$
$$\sum\nolimits = \frac{1}{n+1} - (\frac{1}{n+2}-\frac{1}{n+3}) - {...} -\frac{1}{n+p} < \frac{1}{n+1} < \frac{1}{N} < \epsilon$$
2. $n$ even, $p$ odd: (implying both $n+1$ and $n+p$ are odd)
$$\sum\nolimits = (\frac{1}{n+1}-\frac{1}{n+2}) + {...} + (\frac{1}{n+p-2}-\frac{1}{n+p-1})+\frac{1}{n+p} > 0 > -\epsilon$$
$$\sum\nolimits = \frac{1}{n+1} - (\frac{1}{n+2}-\frac{1}{n+3}) - {...} - (\frac{1}{n+p-1}-\frac{1}{n+p}) < \frac{1}{n+1} < \frac{1}{N} < \epsilon$$
Other cases are left as exercise to OP, the presentation is becoming a bit tedious here.
