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  "body": "\n## Proof\n\nWe add the exponential series of $iz$ and $-iz$.\n\n1. Fix $z \\in \\C$. The series $\\sum_{m \\ge 0} \\abs{z}^m / m!$ converges, since the series of the [[complex exponential|054VXHMCA3]] converges absolutely at $z$.\n2. The partial sums of $\\sum_{n \\ge 0} \\abs{z}^{2n}/(2n)!$ are at most $\\sum_{m \\ge 0} \\abs{z}^m / m!$, because they only omit nonnegative terms. A [[series of nonnegative terms with bounded partial sums converges|0MX1HZBBXJ]]. So $\\sum_{n \\ge 0} (-1)^n z^{2n}/(2n)!$ converges absolutely.\n3. By definition, $e^{iz} = \\sum_{m \\ge 0} (iz)^m/m!$ and $e^{-iz} = \\sum_{m \\ge 0} (-iz)^m/m!$. Both series converge.\n4. The sum of two convergent series converges to the sum of their sums. So $e^{iz} + e^{-iz} = \\sum_{m \\ge 0} c_m z^m / m!$ with $c_m = i^m + (-i)^m$.\n5. Since $(-i)^m = (-1)^m i^m$, we have $c_m = i^m (1 + (-1)^m)$. So $c_m = 0$ for odd $m$. For $m = 2n$, $c_m = 2 i^{2n} = 2(-1)^n$.\n6. Let $S_M$ be the partial sum of the series in step 4 up to $m = M$. By step 5, $S_{2N+1} = \\sum_{n=0}^{N} 2 (-1)^n z^{2n}/(2n)!$.\n7. The sequence $(S_M)$ converges to $e^{iz} + e^{-iz}$, so its subsequence $(S_{2N+1})$ has the same limit. Hence $\\sum_{n=0}^{\\infty} 2(-1)^n z^{2n}/(2n)! = e^{iz} + e^{-iz}$.\n8. Dividing by $2$ and using the definition of the [[complex cosine|0A7YDKRNR6]], $\\cos z = \\sum_{n=0}^{\\infty} (-1)^n z^{2n}/(2n)!$.\n\n## Idea\n\nIn $e^{iz} + e^{-iz}$ the odd powers of $z$ come with $i^m$ and $-i^m$, so they cancel. The even powers double, and $i^{2n} = (-1)^n$ gives the alternating signs.\n",
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