Proving convergence:
$$\sum_{n=1}^\infty (-1)^{n-1}\frac1n$$
Just wanted to confirm if the reason they converge is due to the fact that for n= 1, 3, 5, ... we have a positive harmonic series and for n= 2, 4, 6, ... we have a negative harmonic series? therefore the two cancel giving a convergent sum.
Does the above proof make mathematical sense?
Is it possible for the difference of two particular (not the same) harmonic series to be divergent?
The two do not cancel at all. Perhaps you want to read about Leibniz Alternating Series
About your last question: for any $\,k,m\in\Bbb R\,$ , the series
$$\sum_{n=1}^\infty\left(\frac{1}{kn}-\frac{1}{mn}\right)=\frac{m-k}{mk}\sum_{n=1}^\infty\frac{1}{n}\,\,\,\text{diverges}$$
Second question:
The difference of two complete harmonic series converges to $log(\frac{p}{q})$, where $p$ positive terms and $q$ negative terms are taken every $p+q$ terms. $$ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$
See this message for some examples https://math.stackexchange.com/a/1602987/134791
We can think those pairs as the same series in the sense that they have the same terms but also as different series because their relative speed or rate is not the same. One may be interpreted to run faster than the other, because has more terms than the other per term in the difference.
Here I'm speaking about the series $$\sum_{n=1}^{\infty}(-1)^n\frac1n$$ (I assume this is what you intended, because $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\frac1n}=\sum_{n=1}^{\infty}(-1)^nn$ manifestly can't converge.)
Well, yes, the fact that for $n=1,3,5,\dots$ ($n=2,4,6,\dots$) you have a positive (negative) contribution to the series is the reason why it converges, but that's just an observation and definitely not a proof.
A demonstration easily follows from Leibniz criterion for alternating series, since the sequence $(a_n=\frac1n)$ is monotone decreasing and $\displaystyle \lim_{n\rightarrow\infty}a_n=0$.
