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  "body": "\n## Proof\n\n1. Let $s_N = \\sum_{n=0}^{N} a_n$ be the [[partial sums|0NB643K60Q]]. Since the series is a [[convergent series|0CVVKY7GY2]], $s_N \\to S$ for a real number $S$, the [[sum of the series|0K0H5E3Z5N]].\n2. For each $n \\ge 1$, the partial sums satisfy $a_n = s_n - s_{n-1}$.\n3. Let $\\varepsilon \u003e 0$. By the definition of the [[limit|0XKWSSY8P2]], there is $N$ with $\\abs{s_m - S} \u003c \\varepsilon / 2$ for all $m \\ge N$.\n4. Let $n \\ge N + 1$. Then both $n \\ge N$ and $n - 1 \\ge N$.\n5. By step 2 and the [[triangle inequality|03HF5PMY5T]], $\\abs{a_n} = \\abs{(s_n - S) + (S - s_{n-1})} \\le \\abs{s_n - S} + \\abs{s_{n-1} - S}$.\n6. By steps 3 and 4, the right side of step 5 is less than $\\varepsilon/2 + \\varepsilon/2 = \\varepsilon$.\n7. So $\\abs{a_n - 0} \u003c \\varepsilon$ for all $n \\ge N + 1$. Since $\\varepsilon \u003e 0$ was arbitrary, $a_n \\to 0$.\n8. The second form is the contrapositive of the first. If $(a_n)$ does not converge to $0$, the series cannot converge, by steps 1 to 7. So it diverges.\n\n## Idea\n\nEach term is the difference of two consecutive partial sums. When the partial sums settle down to a limit, consecutive ones get close to each other, so their difference tends to $0$.\n",
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