changing the signs of the harmonic series so that it converges? I need to show that the harmonic series $1+{1\over2}+{1\over3}+\cdots$ remains divergent, when the signs are so changed that we have throughout alternately $p$ positive terms and $q$ negative ones, with $p \ne q$. If $p=q$, however, the resulting series is convergent (it's assumed both $p$ and $q$ finite). The proof given here Change of signs in harmonic series makes use of Abel Formula, which I don't have studied yet, so there must be a different proof based on $H_n - \log n \to \gamma$. The problem is that I cannot bound the partial sums using this formula. We can study the general term of the series obtained by grouping each $p+q$ consecutive terms maybe, for then it follows that the original series converges or diverges with this one. This general term however is hard to study, and the resulting series even more.
For anyone who needs a reference, this exercise is taken from Knopp's book.
 A: Write $r=p+q$, $S_n$ for the partial sum of your series to the $n$th term. Then
$$ S_{(n+1)r} = S_{nr} + (H_{nr+p}-H_{nr}) - (H_{(n+1)r}-H_{nr+p}) = S_{nr} - H_{nr} + 2H_{nr+p} - H_{(n+1)r} \\
  $$
Then for large $n$,
$$ S_{(n+1)r}-S_{nr} \sim - \log nr + 2\log (nr+p)  - \log (n+1)r = \log\left( \frac{(nr+p)^2}{n(n+1)r^2} \right) . $$
The numerator is $ (nr+p)((n+1)r-q) = n(n+1)r^2 + nr(p-q) + p^2 $, so
$$ \log\left( \frac{(nr+p)^2}{n(n+1)r^2} \right) = \log\left( 1 + \frac{r(p-q)}{n+1} + \frac{p^2}{n(n+1)r^2} \right) $$
If $p=q$, a standard logarithm inequality implies this converges, and since the individual terms converge to $0$, this is sufficient for the whole series to converge. On the other hand, if not, e.g. a careful application of some other inequality for the logarithm shows that the difference is bounded below by a multiple of $1/n$, and so $S_{nr}$ diverges, so the series cannot converge. (One could also use the limit comparison test.)
A: Proof sketch: If $p = q$, then the sum of each $p+q$ part is comparable in size to the square of the first term of the part (for instance, if $p = q = 1$, we have $\frac1n - \frac1{n+1} = \frac1{n(n+1)}< \frac1{n^2}$). And $\sum \frac{1}{n^2}$ converges.
However, if $p\neq q$, the "extra" $|p-q|$ terms are comparable in size to the first term, and thus the sum of those terms diverges (for instance, if $p = 2, q = 1$, we have $\frac{1}{n} + \frac1{n+1} - \frac1{n+2} > \frac1n$).
