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. 

The approach I would use would be to expand the sequences and see whether or not a lower or upper limit can be placed.
For $\frac1k$, consider that:
$$\frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + ...$$
...is certainly a larger sum than:
$$\frac12 + \frac14 + \frac14 + \frac18 + \frac18 + \frac18 + \frac18 + \frac1{16} + \frac1{16} + \frac1{16} + \frac1{16} + ...$$
...or:
$$\frac12 + \frac12 + \frac12 + ...$$
So think of it this way.  No matter how many $\frac12$'s you want to add together, you'll eventually get there.  If you want $n \frac12$'s to add together, you will get there by (around) the ($2^n$)'th term.  So any number you come up with can be topped by going further into the series, which means the series diverges to infinity: you can't place any finite upper bound on it that won't be surpassed.
For $ ± \frac1k$ with the $-1$'s alternating, again we look at the partially expanded series:
$$1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} + ...$$
If we combine each pair of terms, we get the series:
$$\frac12 + \frac1{12} + \frac1{30} + \frac1{56} + ...$$
This is clearly bounded below because every term is positive.  To show convergence we have to find some finite upper limit this can't ever exceed.  Doing this directly is tricky, but we can reduce it to a common problem we already know the answer to.  The series is equivalent to:
$$\frac1{1*2} + \frac1{3*4} + \frac1{5*6} + \frac1{7*8} + ...$$
...which is obviously less than:
$$\frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + ...$$
...which, you probably already know, converges to a finite value.
The moral of all this writing I'm doing is that you should try some basic mathematical logic first to see where it can take you.  None of what I did above required any sort of formula (with the exception of the very last step which relies on a common proof you can look up).  It just required me to start expanding the series, group terms, and see what can happen.  You have to be careful when doing this with an infinite series, but the logic holds for the above examples.
The best part about being able to do this logically as opposed to formulaically is that logic can be applied to any problem of this sort, while formulas don't always work or cover what you need--and plus, why use a formula if you don't understand why that formula works in the first place?  This is common in computer science in the analyzation of recursive running times: using the "master method" formula won't work in all cases, but expanding out the series and using logic will always work.
