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  "body": "\n## Proof\n\nLet $x_n = \\abs{a_n}^{1/n}$ for $n \\ge 1$. The term $a_0$ is a single term, and it does not affect convergence. Each $x_n \\ge 0$, and $\\abs{a_n} = x_n^n$.\nThe value $C = +\\infty$ means that $(x_n)$ is not bounded above. Otherwise $(x_n)$ is a [[bounded sequence|05FQ3VGAR9]], and by the definition of the [[limit superior|0P8BA7T7S5]], $C = \\inf_{N \\ge 1} u_N$ with $u_N = \\sup_{k \\ge N} x_k$.\n\n1. Assume $C \u003c 1$. Then $C$ is a real number, and $C \\ge 0$ because every $x_k \\ge 0$. Let $r = (C + 1)/2$, so $C \u003c r \u003c 1$ and $r \u003e 0$.\n2. Since $C = \\inf_N u_N \u003c r$, there is $N \\ge 1$ with $u_N \u003c r$. So $x_k \u003c r$ for every $k \\ge N$.\n3. Then $\\abs{a_k} = x_k^k \u003c r^k$ for $k \\ge N$.\n4. Let $M \\ge N$. By step 3 and the [[sum of a finite geometric series|05APWSX53S]], $\\sum_{k=N}^{M} \\abs{a_k} \\le \\sum_{k=0}^{M} r^k = \\frac{1 - r^{M+1}}{1 - r} \\le \\frac{1}{1 - r}$.\n5. So every partial sum of $\\sum \\abs{a_n}$ is at most $\\sum_{k=0}^{N-1} \\abs{a_k} + 1/(1-r)$.\n6. By the [[criterion for nonnegative series|0MX1HZBBXJ]], $\\sum \\abs{a_n}$ converges. So $\\sum a_n$ converges [[absolutely|0W3K3QPEWT]].\n7. Now assume $C \u003e 1$. We claim that for each $N \\ge 1$ there is $k \\ge N$ with $x_k \u003e 1$.\n8. If $C = +\\infty$, then $(x_n)$ is not bounded above. If $x_k \\le 1$ for all $k \\ge N$, then $\\max(x_1, \\dots, x_{N-1}, 1)$ would bound $(x_n)$ above. So some $k \\ge N$ has $x_k \u003e 1$.\n9. If $C$ is a real number, then $u_N \\ge C \u003e 1$, since $C$ is the infimum of the $u_N$. By the [[approximation property of the supremum|0M8TGGP9JP]] with $c = 1 \u003c u_N$, some $k \\ge N$ has $x_k \u003e 1$.\n10. By steps 7 to 9, there are infinitely many $k$ with $x_k \u003e 1$, and so with $\\abs{a_k} = x_k^k \u003e 1$.\n11. Hence $(a_n)$ does not tend to $0$: no tail of the sequence stays in the interval $(-1, 1)$.\n12. By the [[term test|0VVY4H92G2]], $\\sum a_n$ diverges.\n\n## Idea\n\nIf the $n$-th roots of $\\abs{a_n}$ are eventually below some $r \u003c 1$, then $\\abs{a_n} \u003c r^n$ and a geometric series dominates. If they exceed $1$ infinitely often, then infinitely many terms have absolute value above $1$, so the terms cannot tend to $0$.\n",
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