Suppose that for $0 \leq r \leq 2^n-1$ and $n \geq 0$, we have $$a_{2^n+r}=(-1)^n \frac{1}{2^n+r}.$$ Then we need to show that the series $$\sum_{k=1}^\infty a_k = 1-\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}-\frac{1}{8}-\dots-\frac{1}{15}+\frac{1}{16}+\dots$$ diverges.
The way by which I have done this is to use the following (which can be checked essentially by using an integral test) $$\frac{1}{k}+\frac{1}{k+1}+\dots + \frac{1}{2k-1} \to \log 2^{+} \ \ \ (\text{as} \ k \to \infty)$$ and so $$\frac{1}{2^n} + \frac{1}{2^n+1}+\dots+\frac{1}{2^{n+1}-1} \to \log 2^{+}.$$ Since these expressions tend to $\log 2$ from above, they always have a value greater than $\log 2$.
Then if we consider the $(1+2+4+\dots+2^m)$th partial sum for some $m \in \mathbb N$, then this differs from the $(1+2+4+\dots+2^m + 2^{m+1})$th partial sum by at least $\log 2$ and hence the $k$th partial sums of the original series do not form a Cauchy sequence, and hence it must diverge.
Are there any alternative, potentially more direct ways, of showing that the series diverges?