With $a_k/b_k \to 0$, convergence of $\sum a_k$ does not give convergence of $\sum b_k$

Refuted claim: if $a_k, b_k > 0$ and $(a_k/b_k)$ converges with $\lim_k a_k/b_k = 0$, then convergence of $\sum a_k$ implies convergence of $\sum b_k$.

Claim 2 of For $a_k, b_k > 0$ with $a_k/b_k \to L$: if $L \in (0,\infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each gives the implication in the other direction only: at $L = 0$, convergence of $\sum b_k$ gives convergence of $\sum a_k$. The claim above reverses it, and the reversal fails. Take

$a_k := \frac{1}{(k+1)^{2}}, \qquad b_k := \frac{1}{k+1} \qquad (k \in \mathbb{N}) .$

Both are positive, and $a_k/b_k = 1/(k+1) \to 0$. But $\sum a_k$ is $\sum_{k \ge 1} 1/k^{2}$, which converges (For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ at $p = 2$), while $\sum b_k$ is the harmonic series, which diverges (The harmonic series $\sum 1/k$ diverges, by condensation and by Oresme block grouping).

The asymmetry is not an artefact of the proof. At $L = 0$ the hypothesis says the $a_k$ are eventually much smaller than the $b_k$; smallness of the $a_k$ can never constrain the $b_k$ from above, and the witness shows that it does not.

• The same pair also shows the divergence form is one-directional. Read contrapositively, claim 2 says divergence of $\sum a_k$ forces divergence of $\sum b_k$. The witness has $\sum b_k$ divergent and $\sum a_k$ convergent, so divergence of the larger series says nothing about the smaller one, which is the same asymmetry seen from the other side.

• The regime $L = +\infty$ fails symmetrically. Exchanging the roles of $(a_k)$ and $(b_k)$ in the witness gives $b_k/a_k \to +\infty$ with $\sum b_k$ divergent and $\sum a_k$ convergent, so claim 3 of the test is one-directional for the same reason. That reading is immediate from the computation above, the two sequences being the same two.