Convergence of "alternating" harmonic series where sign is +, --, +++, ----, etc. Exercise 11 from section 9.3 of Introduction to Real Analysis (Bartle):

Can Dirichlet’s Test be applied to establish the convergence of
  $$ 1 - \dfrac12 - \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 - \cdots $$
  $\qquad \qquad$    where the number of signs increases by one in each ‘‘block’’? If not, use another method to 
  establish the convergence of this series.

Dirichlet's test cannot be used because the partial sums generated by (1, -1, -1, 1, 1, 1, ...) are not bounded. But we can group the terms of the series in the following way:
$$ 1 - \left(\dfrac12 + \dfrac13\right) + \left(\dfrac14 + \dfrac15 + \dfrac16\right) - \left( \dfrac17 + \dfrac18 + \dfrac19 + \dfrac{1}{10} \right) + \cdots \\ = \sum _{n=1}^{\infty}(-1)^{n+1}a_n $$
where
$$ (a_n) = \left(1, \left(\dfrac12 + \dfrac13\right), \left(\dfrac14 + \dfrac15 + \dfrac16\right), ... \right) $$
So by Leibniz's test, if the sequence $(a_n)$ is decreasing and $\lim{a_n} = 0$ then the grouped series is convergent. I've shown that since we are grouping terms of the same sign it is sufficient to show the convergence of the grouped series. I've shown that $\lim{a_n} = 0$, but how do I show that $(a_n)$ is decreasing?
 A: Although Dirichlet's test per se does not apply, it seems like a Good Thing to note that the proof of Dirichlet does apply. Sum by parts and you're done.
In more detail: Let $(\epsilon_j)$ be the sequence of plus and minus ones, so we're considering convergence of $$\sum_j\frac{\epsilon_j}{j}.$$Let $\sigma_n=\sum_{j=1}^n\epsilon_j$. Dirchlet does not apply because $\sigma_n$ is not bounded. But it's not hard to see that $$|\sigma_n|\le c\sqrt n.$$(After $N$ "blocks" of ones and minus ones we have $|\sigma_n|\le c N$. But after $N$ blocks we have $n\sim N^2$.)
So sum by parts https://en.wikipedia.org/wiki/Summation_by_parts :
$$\sum_{j=1}^n\frac{\epsilon_j}{j}=\frac{\sigma_n}{n}-\sum_{j=1}^{n-1}\sigma_j
\left(\frac1{j+1}-\frac1j\right).$$Since $\sqrt n/n\to0$ and $\sum\sqrt j/j^2<\infty$ the sum converges.
Moral Proofs of theorems are even better than theorems.
A: Note that $a_n = \sum_{k=n(n-1)/2+1}^{n(n+1)/2} \frac1k$. In particular, since $\frac1x$ is decreasing,
$$
\int_{n(n-1)/2+1}^{n(n+1)/2+1} \frac{dx}x < a_n < \int_{n(n-1)/2}^{n(n+1)/2} \frac{dx}x,
$$
or
$$
\log\frac{n^2+n+2}{n^2-n+2} < a_n < \log\frac{n+1}{n-1}.
$$
In particular,
$$
a_n-a_{n+1} > \log\frac{n^2+n+2}{n^2-n+2} - \log\frac{n+2}n = \log\bigg( 1+\frac{2(n-2)}{n^3+n^2+4} \bigg) \ge0
$$
for $n\ge2$.
(In fact, the estimate $|a_{2n-1}-a_{2n}|<\frac1{n^2}$ would suffice to establish convergence, regardless of whether the $a_n$ are decreasing.)
