Convergence of "alternating" harmonic series where sign is +, --, +++, ----, etc.

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.)

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.