"Altered" Alternating Series Diverges or Converges? Consider the series
$$
1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
which alternates between a block of positives and a block of negatives, with the block sizes, 
$$
1, 1, 2, 3, 4, 5, 6, \dots
$$
and so on.
Does this series converge or diverge?

Write $H_n = \sum_{k=1}^{n} \frac{1}{k} $ for the $n$-th harmonic number. It is well-known that $H_n$ has the asymptotic form
$$ H_n = \log n + \gamma + \mathcal{O}\left(\frac{1}{n}\right) \quad \text{as} \quad n\to\infty. $$
Now returning to OP's question, let $s_n$ denote the $n$-th partial sum. Also, write
$$m_k = \frac{k(k+1)}{2} + 1 \qquad \text{and} \qquad K_n = \max\{ k \geq 1 : m_k \leq n \}.$$
Using this, we can write $s_n$ in the the following block form
$$ s_n = 1 + \Bigg( \sum_{k = 1}^{K_n} (-1)^k (H_{m_k} - H_{m_{k-1}}) \Bigg) + (-1)^{K_n+1} (H_n - H_{m_{K_n}} ). $$
Then by using the asymptotic form of $H_n$, we easily find that
$$ H_{m_k} - H_{m_{k-1}} = \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right), $$
and so,
$$ s_n = 1 + \sum_{k=1}^{K_n} \left( (-1)^{k} \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right) \right) + \mathcal{O}\left(\frac{1}{K_n}\right). $$
Therefore it follows that $s_n$ converges as $n\to\infty$
