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

Let $(a_k)$ and $(b_k)$ be sequences of reals with $a_k > 0$ and $b_k > 0$ for every $k \in \mathbb{N}$, and put $q_k := a_k / b_k$. Then:

1. if $(q_k)$ converges with $\lim_k q_k = L$ for a real $L > 0$ (Limits and Cauchy sequences of reals), then $\sum a_k$ converges if and only if $\sum b_k$ converges;
2. if $(q_k)$ converges with $\lim_k q_k = 0$, then convergence of $\sum b_k$ implies convergence of $\sum a_k$; equivalently, divergence of $\sum a_k$ implies divergence of $\sum b_k$;
3. if $(q_k)$ diverges to $+\infty$ (Divergence to $+\infty$ and to $-\infty$), then convergence of $\sum a_k$ implies convergence of $\sum b_k$; equivalently, divergence of $\sum b_k$ implies divergence of $\sum a_k$.

In each clause the convergence of $(q_k)$, or its divergence to $+\infty$, is part of the hypothesis, so the symbol $\lim_k q_k$ denotes wherever it is written (A sequence has at most one limit).

Neither implication in claim 2 can be reversed, and by symmetry neither can the one in claim 3; the companion page exhibits a pair with $\lim_k q_k = 0$, $\sum a_k$ convergent and $\sum b_k$ divergent.

For families from a general starting index $m$ the statement is the same, applied to the shifted sequences $j \mapsto a_{j+m}$ and $j \mapsto b_{j+m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

On the third regime. "$L = \infty$" is written here as divergence of $(q_k)$ to $+\infty$ in the sense of Divergence to $+\infty$ and to $-\infty$, and never as a limit equation with an infinite right-hand side. A sequence diverging to $+\infty$ has no limit in $\mathbb{R}$, and this library does not write $\lim q_k = +\infty$.

• Why the three regimes are treated as one proof. The Statement is a conjunction of three implications, each with its own hypothesis on $(q_k)$. Fixing the two sequences and arguing by cases on which regime holds proves all three at once, and costs nothing: if none of the regimes holds, every one of the three implications is vacuously true.

• Positivity of $(b_k)$ is needed twice. It is what makes $q_k$ defined at all, and it is what lets an inequality between the $q_k$ be multiplied through to an inequality between the $a_k$ and the $b_k$ without reversing. Positivity of $(a_k)$ is what supplies the lower bound $0$ that the direct comparison test requires.

• The limit is only used through an eventual two-sided estimate. No step needs the exact value of $L$, only that $q_k$ is eventually trapped strictly between two positive multiples of it. That is why the test still works when the quotients merely stay between two positive constants, and why the hypothesis $\lim_k q_k = L$ is stronger than what the proof consumes.