[proofplan] We prove absolute convergence for $L < 1$ by placing the tail of $(|a_j|)$ below a convergent geometric sequence. We prove divergence for $L > 1$ by showing that the terms $a_j$ cannot tend to $0$, so the necessary condition for [series](/pages/1364) convergence fails. Finally, we prove that the case $L = 1$ is inconclusive by giving one divergent and one absolutely convergent series for which the same root [limit](/pages/1146) equals $1$. [/proofplan] [step:Compare the tail with a convergent geometric series when $L < 1$] Assume $L < 1$. Choose a real number $r$ such that $L < r < 1$. Since $|a_j|^{1/j} \to L$, the definition of [convergence of a sequence](/pages/1211) gives an integer $N \in \mathbb{N}$ such that \begin{align*} |a_j|^{1/j} < r \quad \text{for all } j \geq N. \end{align*} Because $j \in \mathbb{N}$ and both sides are non-negative, raising both sides to the $j$-th power preserves the inequality: \begin{align*} |a_j| < r^j \quad \text{for all } j \geq N. \end{align*} The geometric series $\sum_{j=N}^{\infty} r^j$ converges because $0 < r < 1$. Hence, by the [Comparison Test](/theorems/173) applied to the non-negative sequences $(|a_j|)_{j=N}^{\infty}$ and $(r^j)_{j=N}^{\infty}$, the series $\sum_{j=N}^{\infty} |a_j|$ converges. Adding the finite sum $\sum_{j=1}^{N-1} |a_j|$ preserves convergence, so $\sum_{j=1}^{\infty} |a_j|$ converges. Therefore $\sum_{j=1}^{\infty} a_j$ converges absolutely. [guided] Assume $L < 1$. The goal is to dominate the absolute values $|a_j|$ by a geometric sequence whose ratio is strictly less than $1$. Since $L$ is strictly below $1$, choose a real number $r$ between them: \begin{align*} L < r < 1. \end{align*} The hypothesis $|a_j|^{1/j} \to L$ means that the sequence of roots is eventually as close to $L$ as we want. In particular, because $r$ is strictly larger than $L$, there exists $N \in \mathbb{N}$ such that \begin{align*} |a_j|^{1/j} < r \quad \text{for all } j \geq N. \end{align*} For each $j \geq N$, both $|a_j|^{1/j}$ and $r$ are non-negative, and $j$ is a positive integer. Therefore raising the inequality to the $j$-th power is order-preserving: \begin{align*} |a_j| < r^j \quad \text{for all } j \geq N. \end{align*} Now $\sum_{j=N}^{\infty} r^j$ is a convergent geometric series because its ratio satisfies $0 < r < 1$. The Comparison Test applies to the non-negative tail terms because \begin{align*} 0 \leq |a_j| < r^j \quad \text{for all } j \geq N. \end{align*} Thus $\sum_{j=N}^{\infty} |a_j|$ converges. A finite number of initial terms cannot affect convergence of a series, so adding $\sum_{j=1}^{N-1} |a_j|$ gives convergence of $\sum_{j=1}^{\infty} |a_j|$. This is exactly absolute convergence of $\sum_{j=1}^{\infty} a_j$. [/guided] [/step] [step:Force the terms away from zero when $L > 1$] Assume $L > 1$. Since $|a_j|^{1/j} \to L$, the definition of convergence gives an integer $N \in \mathbb{N}$ such that \begin{align*} |a_j|^{1/j} > 1 \quad \text{for all } j \geq N. \end{align*} Raising both sides to the $j$-th power gives \begin{align*} |a_j| > 1 \quad \text{for all } j \geq N. \end{align*} Hence $a_j$ does not converge to $0$. By the Divergence Test, the series $\sum_{j=1}^{\infty} a_j$ diverges. [/step] [step:Exhibit both convergence and divergence when the root limit equals $1$] Define two real sequences $b, c: \mathbb{N} \to \mathbb{R}$ by \begin{align*} b(j) &= \frac{1}{j}, & c(j) &= \frac{1}{j^2}. \end{align*} For $b_j := b(j)$, the harmonic series $\sum_{j=1}^{\infty} b_j = \sum_{j=1}^{\infty} \frac{1}{j}$ diverges. For $c_j := c(j)$, the $p$-series $\sum_{j=1}^{\infty} c_j = \sum_{j=1}^{\infty} \frac{1}{j^2}$ converges. It remains to check that both examples have root limit $1$. Since $\lim_{j \to \infty} j^{1/j} = 1$, we have \begin{align*} \lim_{j \to \infty} |b_j|^{1/j} &= \lim_{j \to \infty} \left(\frac{1}{j}\right)^{1/j} = \lim_{j \to \infty} \frac{1}{j^{1/j}} = 1, \\ \lim_{j \to \infty} |c_j|^{1/j} &= \lim_{j \to \infty} \left(\frac{1}{j^2}\right)^{1/j} = \lim_{j \to \infty} \frac{1}{(j^{1/j})^2} = 1. \end{align*} Thus the same value $L = 1$ occurs for a divergent series and for a convergent series, so the root test gives no conclusion when $L = 1$. [/step]