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  "body": "\n## Proof\n\n1. Apply the definition of the [[limit|0XKWSSY8P2]] with $\\varepsilon = c/2 \u003e 0$. There is $N$ such that $\\abs{a_n/b_n - c} \u003c c/2$ for all $n \\ge N$.\n2. So $c/2 \u003c a_n / b_n \u003c 3c/2$ for $n \\ge N$. Since $b_n \u003e 0$, this gives $a_n \u003c \\tfrac{3c}{2} b_n$ and $b_n \u003c \\tfrac{2}{c} a_n$ for $n \\ge N$.\n3. Both series have nonnegative terms. By the [[criterion for nonnegative series|0MX1HZBBXJ]], each converges if and only if its [[partial sums|0NB643K60Q]] are bounded above.\n4. Assume that $\\sum b_n$ converges, and let $B$ be its sum. Since $b_n \\ge 0$, every finite sum of terms $b_n$ is at most $B$. This follows from the same criterion, as in the [[comparison test|04Z5G0THN6]].\n5. For $M \\ge N$, step 2 gives\n$$\\sum_{n=0}^{M} a_n \\le \\sum_{n=0}^{N-1} a_n + \\frac{3c}{2} \\sum_{n=N}^{M} b_n \\le \\sum_{n=0}^{N-1} a_n + \\frac{3c}{2} B.$$\n6. The right side of step 5 does not depend on $M$. For $M \u003c N$, the partial sum is at most $\\sum_{n=0}^{N-1} a_n$. So the partial sums of $\\sum a_n$ are bounded above, and $\\sum a_n$ converges by step 3.\n7. Conversely, assume that $\\sum a_n$ converges, with sum $A$. In the same way, for $M \\ge N$,\n$$\\sum_{n=0}^{M} b_n \\le \\sum_{n=0}^{N-1} b_n + \\frac{2}{c} A.$$\n8. So the partial sums of $\\sum b_n$ are bounded above, and $\\sum b_n$ converges by step 3.\n\n## Idea\n\nWhen $a_n / b_n$ tends to a positive number $c$, the two sequences are eventually within constant factors of each other. Constant factors and finitely many initial terms cannot change whether partial sums stay bounded.\n",
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