Convergence of $1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\ldots$

In blocks of $3$ terms, the series is $$ \sum_{n=0}^{\infty} \left(\dfrac{1}{4n+1}+\dfrac{1}{4n+3}-\dfrac{1}{2n+2}\right) = \sum_{n=0}^{\infty} \dfrac{8 n + 5}{32 n^3 + 64 n^2 + 38 n + 6} \le \dfrac56+ \sum_{n=1}^{\infty} \dfrac{1}{n^2} < \infty $$

This proves that the partial sums $S_{3n+2}$ of the original series converge. The other partial sums, $S_{3n+1}$ and $S_{3n}$, differ from $S_{3n+2}$ by one or two terms that converge to zero and so they converge as well, to the same limit. Thus, the partial sums converge, that is, the series converges.

It seems like the series is alternatingly composed of positive terms (and it is okay to group them) of the form

$$a_{2n}=\frac1{4n+1}+\frac1{4n+3}=\frac{8n+4}{(4n+1)(4n+3)}=\frac{8n+4}{16n^2+16n+3}=\frac{n+1/2}{2n^2+2n+3/8}.$$

for $n=0,1,2,...$, and negative terms of the form

$$a_{2n+1}=-\frac{1}{2n+2}.$$

also for $n=0,1,2,...$. Note that for sufficiently large $n$ we have

$$ \underbrace{\frac1{2n}}_{|a_{2n-1}|} \ge \underbrace{\frac{n+1/2}{2n^2+2n+3/8}}_{|a_{2n}|} \ge \underbrace{\frac 1{2n+2}}_{|a_{2n+1}|}. $$

So we have an alternating series with absolutely decreasing terms with $|a_n|\to 0$. We can apply the alternating series test to reason that the sum converges.

As many people already pointed out, the series converges thanks to Leibniz criterion! Indeed the series is $$ \sum_{n=1}^\infty (-1)^{n+1} a_n, $$ where $a_n$ are $$ a_n = \begin{cases} \frac{1}{n} & n\ \text{even,}\\ \frac{1}{n+n-1}+ \frac{1}{n+2+n-1} = \frac{1}{2n-1}+\frac{1}{2n+1} & n \ \text{odd,} \end{cases} \qquad a_n \to 0. $$