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. 

We do not answer the specific question, but instead look at two related series,
$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots\tag{1}$$
and
$$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\cdots\tag{2}$$
Series (1) diverges. For the sum of the first $3$ terms is $2$, the sum of the first $7$ is $3$, the sum of the first $15$ is $4$, the sum of the first $31$ is $5$, and so on. The partial sums "blow up," albeit with a great deal of reluctance. It takes an awful lot of terms to get a sum of $100$.
Series (2) converges. The partial sums are $1, \frac{1}{2}, 1, \frac{3}{4}, 1, \frac{3}{4}, 1,\frac{7}{8}, 1,\frac{7}{8}$, and so on. It is clear that the partial sums approach $1$.
