= Solution
Let $S_N=\sum_{k=1}^Na_k$, with $S_0=0$. The indices in the $n$th dyadic block have one sign, so
$$
\left|S_{2^{n+1}-1}-S_{2^n-1}\right|
=\sum_{r=0}^{2^n-1}\frac1{2^n+r}>\frac{2^n}{2^{n+1}}=\frac12.
$$
A convergent <sequence> of <partial sums> must be a <Cauchy sequence>: for every $\varepsilon>0$, all sufficiently late pairs of <partial sums> differ by less than $\varepsilon$. The displayed differences violate this necessary condition with $\varepsilon=1/4$, at arbitrarily late endpoints. Thus \b[the <series> diverges], even though its individual terms tend to zero. This <dyadic-block alternation of reciprocal terms> has growing block lengths; the fixed-length block argument of the previous part does not apply.
Back to article page
