Prove that the following series diverges I want to prove that the series
$$
\frac{1}{z} + \frac{1}{z+1} + \cdots + \frac{1}{z+p-1} - \frac{1}{z+p} - \frac{1}{z+p+1} - \cdots - \frac{1}{z+2p+q-1} + \frac{1}{z+2p+q} + \cdots, 
$$
in which $(p+q)$ negative terms always follow $p$ positive terms, is divergent.
Note that $z \in \mathbb{C}$ such that the denominator of each term is not $0$. I know it has something to do with the harmonic series but I am not sure how to give a formal proof. I think the hardest part is that $z$ is complex so we cannot use direct comparison test.
This is a question from A Course of Modern Analysisby E.T. Whitaker & G.N. Watson.
 A: Sketch of the proof first consider the series starting at $m=1$ since the $1/z$ term makes no difference (this for simplicity so we can use $z=0$ in what follows):
Note that $|\frac{1}{z+m} - \frac{1}{m}|=\frac{|z|}{m|m+z|} \le \frac{1}{m^2}, m \ge m_K \ge 1$ when $z$ is fixed in a compact set $K$ away from the negative integers since we can take $m_k \ge 2\max_K |z|+1$ for example (in particular we can take the compact set $K_z=${$z$})
This immediately means that the given series (for an allowed fixed $z$) has the same behavior as the series with $z=0$ (starting at $m=m_{z} \ge 1$) and that is obviously divergent if the negative blocks are strictly bigger than the positive blocks by the same amount - to see that, group the $p$ positive terms with the first $p$ negative terms and that gives terms $\frac{1}{m}-\frac{1}{m+p}=\frac{p}{m(m+p)}$ which are absolutely convergent (assuming $p,q \ge 1$ as $p,q$ are fixed), while in every $2p+q$ consecutive block, we remain with $q$ negative terms so that sum is divergent as it is at most $\sum_{n \ge n_k} \frac{-1}{(2q+p)n+p+q}=-\infty$
(edit later)- as asked let me show how one proves the divergence of $1/3-1/4-1/5+1/6-1/7-1/8...$ so more or less the case $p=1, q=1$ where we ignore the first few terms so we can use $0$ as noted; take the partial sums
\begin{align}
s_3 & =1/3-1/4-1/5=(1/3-1/4)-1/5=b_1-1/(6-1) \\[6pt]
s_6 & =(1/3-1/4)+(1/6-1/7)-1/5-1/8 \\[6pt]
& =b_1+b_2-1/(6-1)-1/(9-1),
\end{align}
so
$$s_{3n}=b_1+b_2+\cdots+b_n-\sum_{2 \le k \le n+1}\frac{1}{3k-1}$$
and note that since each partial sum is a sum of finitely many terms we can recombine them as we wish; it is clear that $\sum {b_k}$ is absolutely convergent, so $s_{3n} \to -\infty$, hence the original series is divergent; it is obvious that $s_{3n}-s_{3n \pm 1} \to 0$ so the original series is indeed divergent to $-\infty$
for arbitrary $p,q \ge 1$ we use $s_{(2p+q)n}$ instead and same proof as above; with a little more work, noting that each positive block added from $s_{(2p+q)n+1}$ to $s_{(2p+q)n+p}$ is bounded, we get that $s_m \to -\infty$ so the series diverges to minus infinity in general too
