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  "body": "\n## Example\n\nLet $B_k$ be the $k$-th [[Bernoulli polynomial|BXQKBJJGFG]], and let $k \\ge 2$ be an integer.\nThen for every $x \\in [0, 1]$,\n$$B_k(x) = -\\frac{k!}{(2\\pi i)^k} \\sum_{n \\in \\Z, \\, n \\ne 0} \\frac{e^{2\\pi i n x}}{n^k},$$\nand the series converges absolutely and uniformly on $[0, 1]$.\nSo the $1$-periodic function equal to $B_k$ on $[0, 1)$ has the [[Fourier coefficients|5NN2YZZR3V]] $0$ at $n = 0$ and $-\\frac{k!}{(2\\pi i n)^k}$ at $n \\ne 0$, for period $1$.\n\nCheck of the coefficients: $\\int_0^1 B_k(x) \\, dx = 0$ for $k \\ge 1$, which gives the coefficient at $n = 0$.\nFor $n \\ne 0$, integration by parts with $B_k' = k B_{k-1}$ and $B_k(1) = B_k(0)$ for $k \\ge 2$ gives $\\int_0^1 B_k(x) e^{-2\\pi i n x} \\, dx = \\frac{k}{2\\pi i n} \\int_0^1 B_{k-1}(x) e^{-2\\pi i n x} \\, dx$.\nFor $k = 1$, $\\int_0^1 (x - \\frac{1}{2}) e^{-2\\pi i n x} \\, dx = -\\frac{1}{2\\pi i n}$, and induction gives $-\\frac{k!}{(2\\pi i n)^k}$.\nFor $k \\ge 2$ the coefficients are summable and the periodic extension is continuous, so the series converges uniformly to it.\n\n## Remarks\n\nGrouping $n$ and $-n$ gives real forms; for $k = 2m$ with $m \\ge 1$,\n$$B_{2m}(x) = (-1)^{m+1} \\frac{2 \\, (2m)!}{(2\\pi)^{2m}} \\sum_{n=1}^{\\infty} \\frac{\\cos 2\\pi n x}{n^{2m}}.$$\nAt $x = 0$, where $B_{2m}(0)$ is the Bernoulli number $B_{2m}$, this is Euler's formula $\\zeta(2m) = (-1)^{m+1} \\frac{(2\\pi)^{2m} B_{2m}}{2 \\, (2m)!}$ for the [[Riemann zeta function|1XWBY4N0VT]].\nFor $m = 1$ it gives $\\zeta(2) = \\frac{\\pi^2}{6}$, and for $m = 2$ it gives $\\zeta(4) = \\frac{\\pi^4}{90}$.\nFor $k = 1$ the same series, $-\\sum_{n \\ge 1} \\frac{\\sin 2\\pi n x}{\\pi n}$, converges to $x - \\frac{1}{2}$ only for $0 \u003c x \u003c 1$; at $x = 0$ it gives $0$, the midpoint of the jump.\n\n## Explanations\n\n### L1 human\n\nThe Bernoulli polynomials, repeated periodically, have very simple Fourier coefficients: a constant times $1/n^k$.\nReading the formula at a single point turns it into an exact value of a series such as $\\sum 1/n^2$ or $\\sum 1/n^4$.\nThis is the cleanest route to the values of the zeta function at even integers.\n",
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