"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?
 A: 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 $n\to\infty$
A: Let: $$
S=1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
For a given sequence of additions assume a constant denominator, specifically the smallest in the group. For a series of subtractions pick a common constant denominator, the largest in the group. This guarantees a maximum effect of additions and a minimum effect of subtractions, guaranteeing greater partial sums than the original series. 
Doing this we get:
$$S' = 1 -\frac{1}{2}+\frac{2}{3}-\frac{3}{7}+\frac{4}{8}-\frac{5}{17}+\frac{6}{18}...$$
Suppose the numerator  is $2k$ for some integer $k$. Then the denominator is $\frac{5k^2-5k+6}{2}$. This term is positive. The term directly to the left is negative with numerator $2k-1$ and denominator $\frac{5k^2-5k+4}{2}$.
So:
$$S'=1 + \sum_{k=1}^\infty \frac{4k}{5k^2-5k+6}-\frac{4k-2}{5k^2-5k+4}$$
This preserves the order of terms. 
If the two fractions are added, we get a second degree polynomial in the numerator and a 4th degree polynomial in the denominator for a terms on the order of $O(1/k^2)$. 
So $S'$ converges by comparison to $1/k^2$ and $S$ converges by comparison with $S'$. 
A: Intuitively:
The blocks of constant sign are made of pieces of the Harmonic series, each with $n$ terms, between $\dfrac{n(n-1)}2$ and $\dfrac{n(n+1)}2$. By the asymptotic formula, we known that these sums are asymptotic to $\log\dfrac{n+1}{n-1}\sim\dfrac2n$. Hence the total sum converges, the same way than the ordinary asymptotic series does.
