= Solution
<Group> three consecutive terms. Their block sum is
$$
b_n=-\frac1{2n-1}+\frac1{4n-1}+\frac1{4n}
=-\frac{6n-1}{4n(2n-1)(4n-1)}.
$$
Since $2n-1\ge n$ and $4n-1\ge3n$ for $n\ge1$, we have $|b_n|\le1/(2n^2)$. The <comparison test for series> and the preceding proof of convergence of $\sum n^{-2}$ show that $\sum b_n$ converges absolutely.
This establishes convergence of the original <partial sums> at indices $3N$. Each intermediate <partial sum> differs from one of these by at most two of the original terms; those terms tend to zero. Thus all <partial sums> have the same <limit of a sequence>, which is the mechanism of <convergence from fixed-length series blocks>. The original <series> is not absolutely convergent, since its absolute-value <series> contains $\sum_n(2n-1)^{-1}\ge\tfrac12\sum_n n^{-1}$. Therefore \b[the original <series> converges conditionally].
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