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  "body": "\n## Proof\n\nWe compare $\\abs{a_n}$ with a geometric sequence.\n\n1. Each ratio $\\abs{a_{n+1}}/\\abs{a_n}$ is at least $0$. Since [[limits preserve non-strict inequalities|0N32V0E3XJ]], we get $L \\ge 0$.\n2. Let $r = (L + 1)/2$. Since $0 \\le L \u003c 1$, we have $L \u003c r \u003c 1$ and $r \u003e 0$.\n3. Apply the definition of the [[limit|0XKWSSY8P2]] with $\\varepsilon = r - L \u003e 0$. There is $N$ with $\\abs{a_{n+1}}/\\abs{a_n} \u003c r$ for all $n \\ge N$.\n4. So $\\abs{a_{n+1}} \u003c r \\abs{a_n}$ for $n \\ge N$. By induction on $j$, $\\abs{a_{N+j}} \\le r^j \\abs{a_N}$ for every $j \\ge 0$.\n5. Let $M \\ge N$. By step 4 and the [[sum of a finite geometric series|05APWSX53S]],\n$$\\sum_{n=N}^{M} \\abs{a_n} \\le \\abs{a_N} \\sum_{j=0}^{M-N} r^j = \\abs{a_N} \\frac{1 - r^{M-N+1}}{1 - r} \\le \\frac{\\abs{a_N}}{1 - r}.$$\nThe last inequality uses $0 \u003c r \u003c 1$.\n6. So for every $M$, $\\sum_{n=0}^{M} \\abs{a_n} \\le \\sum_{n=0}^{N-1} \\abs{a_n} + \\abs{a_N}/(1-r)$. This bound does not depend on $M$.\n7. The terms $\\abs{a_n}$ are nonnegative and their partial sums are bounded above. By the [[criterion for nonnegative series|0MX1HZBBXJ]], $\\sum \\abs{a_n}$ converges.\n8. So $\\sum a_n$ converges [[absolutely|0W3K3QPEWT]].\n\n## Idea\n\nIf the ratio of consecutive terms tends to $L \u003c 1$, then eventually each term is at most $r$ times the previous one, for a fixed $r \u003c 1$. So the tail is dominated by a convergent geometric series.\n",
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