[proofplan] The hypothesis says that, eventually, $a_n$ is trapped between two positive constant multiples of $b_n$. We first turn the limit condition into the explicit two-sided estimate $\frac{L}{2}b_n \leq a_n \leq \frac{3L}{2}b_n$ for all sufficiently large $n$. Since changing finitely many terms does not affect convergence of a series, the two-sided estimate compares the tails of the two nonnegative series. Boundedness of the partial sums of one tail then forces boundedness of the partial sums of the other tail, in both directions. [/proofplan] [step:Convert the ratio limit into two-sided tail estimates] Since $L \in (0,\infty)$, define the positive number $\varepsilon := L/2$. By the definition of the limit \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} = L, \end{align*} there exists $N_1 \in \mathbb{N}$ such that for every $n \geq N_1$ the quotient $\frac{a_n}{b_n}$ is defined and satisfies \begin{align*} \left|\frac{a_n}{b_n} - L\right| < \frac{L}{2}. \end{align*} Let $N := \max\{N_0,N_1\}$. Then $b_n > 0$ for every $n \geq N$, and the preceding inequality gives \begin{align*} \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2} \end{align*} for every $n \geq N$. Multiplying by the positive number $b_n$ yields \begin{align*} \frac{L}{2} b_n \leq a_n \leq \frac{3L}{2} b_n \end{align*} for every $n \geq N$. [guided] The purpose of the limit hypothesis is to produce a uniform comparison after finitely many terms. Since the limiting ratio is the strictly positive finite number $L$, we choose the tolerance $\varepsilon := L/2$. The definition of convergence of the numerical sequence $\left(a_n/b_n\right)_{n \geq N_0}$ gives an index $N_1 \in \mathbb{N}$ such that whenever $n \geq N_1$, \begin{align*} \left|\frac{a_n}{b_n} - L\right| < \frac{L}{2}. \end{align*} We also need the quotient to be legitimate, so we keep the index $N_0$ from the hypothesis that $b_n > 0$ eventually. Set $N := \max\{N_0,N_1\}$. Then for every $n \geq N$, both requirements hold: $b_n > 0$ and the ratio is within $L/2$ of $L$. Unpacking the absolute value inequality gives \begin{align*} -\frac{L}{2} < \frac{a_n}{b_n} - L < \frac{L}{2}, \end{align*} hence \begin{align*} \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}. \end{align*} Because $b_n > 0$, multiplying by $b_n$ preserves the inequalities: \begin{align*} \frac{L}{2} b_n \leq a_n \leq \frac{3L}{2} b_n. \end{align*} This is the entire mechanism of the test: after finitely many terms, the two sequences differ only by bounded positive multiplicative constants. [/guided] [/step] [step:Show convergence of $\sum b_n$ implies convergence of $\sum a_n$] Assume $\sum_{n=1}^{\infty} b_n$ converges. Define the partial sums \begin{align*} A_m := \sum_{n=1}^{m} a_n, \qquad B_m := \sum_{n=1}^{m} b_n \end{align*} for $m \in \mathbb{N}$. Since each $a_n \geq 0$, the sequence $(A_m)_{m=1}^{\infty}$ is nondecreasing. Since $\sum_{n=1}^{\infty} b_n$ converges, the sequence $(B_m)_{m=1}^{\infty}$ is bounded above; choose $M_b > 0$ such that $B_m \leq M_b$ for every $m \in \mathbb{N}$. For every $m \geq N$, \begin{align*} A_m &= \sum_{n=1}^{N-1} a_n + \sum_{n=N}^{m} a_n \\ &\leq \sum_{n=1}^{N-1} a_n + \frac{3L}{2}\sum_{n=N}^{m} b_n \\ &\leq \sum_{n=1}^{N-1} a_n + \frac{3L}{2} M_b. \end{align*} Thus $(A_m)_{m=1}^{\infty}$ is bounded above after the index $N$, and the finitely many earlier partial sums are bounded by their maximum. Therefore $(A_m)_{m=1}^{\infty}$ is nondecreasing and bounded above, so it converges. Hence $\sum_{n=1}^{\infty} a_n$ converges. [/step] [step:Show convergence of $\sum a_n$ implies convergence of $\sum b_n$] Assume $\sum_{n=1}^{\infty} a_n$ converges. With the same partial sums \begin{align*} A_m := \sum_{n=1}^{m} a_n, \qquad B_m := \sum_{n=1}^{m} b_n, \end{align*} the sequence $(B_m)_{m=1}^{\infty}$ is nondecreasing because each $b_n \geq 0$. Since $\sum_{n=1}^{\infty} a_n$ converges, the sequence $(A_m)_{m=1}^{\infty}$ is bounded above; choose $M_a > 0$ such that $A_m \leq M_a$ for every $m \in \mathbb{N}$. From the lower bound in the tail estimate, \begin{align*} \frac{L}{2} b_n \leq a_n \end{align*} for every $n \geq N$. Since $L/2 > 0$, this gives \begin{align*} b_n \leq \frac{2}{L}a_n \end{align*} for every $n \geq N$. Therefore, for every $m \geq N$, \begin{align*} B_m &= \sum_{n=1}^{N-1} b_n + \sum_{n=N}^{m} b_n \\ &\leq \sum_{n=1}^{N-1} b_n + \frac{2}{L}\sum_{n=N}^{m} a_n \\ &\leq \sum_{n=1}^{N-1} b_n + \frac{2}{L}M_a. \end{align*} Thus $(B_m)_{m=1}^{\infty}$ is nondecreasing and bounded above, so it converges. Hence $\sum_{n=1}^{\infty} b_n$ converges. [/step] [step:Conclude the equivalence] We have proved both implications: \begin{align*} \sum_{n=1}^{\infty} b_n \text{ converges} \implies \sum_{n=1}^{\infty} a_n \text{ converges} \end{align*} and \begin{align*} \sum_{n=1}^{\infty} a_n \text{ converges} \implies \sum_{n=1}^{\infty} b_n \text{ converges}. \end{align*} Therefore $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^{\infty} b_n$ converges. [/step]