[proofplan] We prove the Ratio Test in two cases. When $L < 1$, we sandwich the ratio below a geometric rate $r \in (L, 1)$ for large indices, compare with the convergent geometric series $\sum r^k$, and apply the [Comparison Test](/theorems/173). When $L > 1$, we show the terms grow geometrically, so $|a_j| \not\to 0$ and the series diverges by the divergence test. [/proofplan] [step:Bound the terms geometrically when $L < 1$ and apply the Comparison Test] Choose $r$ with $L < r < 1$. Since $|a_{j+1}/a_j| \to L < r$, there exists $N \in \mathbb{N}$ such that $|a_{j+1}/a_j| < r$ for all $j \geq N$. Iterating this bound $k$ times starting from index $N$: \begin{align*} |a_{N+k}| < r^k |a_N| \quad \text{for all } k \geq 0. \end{align*} The series $\sum_{k=0}^{\infty} r^k |a_N| = |a_N| / (1 - r)$ is a convergent geometric series since $0 < r < 1$. By the [Comparison Test](/theorems/173) applied to $|a_{N+k}| \leq r^k |a_N|$, the series $\sum_{j=N}^{\infty} |a_j|$ converges. Adding the finitely many initial terms, $\sum_{j=1}^{\infty} |a_j|$ converges. [guided] When $L < 1$, the consecutive ratios $|a_{j+1}/a_j|$ are eventually less than some fixed $r < 1$. This means the terms decay at least as fast as a geometric sequence. Choose $r$ with $L < r < 1$. By the definition of limit, there exists $N \in \mathbb{N}$ such that $|a_{j+1}/a_j| < r$ for all $j \geq N$. Unrolling this inequality: \begin{align*} |a_{N+1}| &< r|a_N|, \\ |a_{N+2}| &< r|a_{N+1}| < r^2|a_N|, \\ &\;\;\vdots \\ |a_{N+k}| &< r^k|a_N|. \end{align*} Since $0 < r < 1$, the geometric series $\sum_{k=0}^{\infty} r^k |a_N|$ converges. The [Comparison Test](/theorems/173) then gives convergence of $\sum_{j=N}^{\infty} |a_j|$. Since adding finitely many terms does not affect convergence, $\sum_{j=1}^{\infty} |a_j|$ converges. [/guided] [/step] [step:Show the terms grow without bound when $L > 1$, forcing divergence] Choose $r$ with $1 < r < L$. There exists $N \in \mathbb{N}$ such that $|a_{j+1}/a_j| > r$ for all $j \geq N$. Iterating: \begin{align*} |a_{N+k}| > r^k |a_N| \quad \text{for all } k \geq 0. \end{align*} Since $r > 1$ and $|a_N| > 0$, we have $r^k |a_N| \to \infty$ as $k \to \infty$. Therefore $|a_j| \not\to 0$, and the series $\sum_{j=1}^{\infty} a_j$ diverges by the divergence test (a necessary condition for convergence of a series is that its terms tend to zero). [/step]