# §24.8 Series Expansions

## §24.8(i) Fourier Series

If $n=1,2,\dots$ and $0\leq x\leq 1$, then

 24.8.1 $\displaystyle\mathop{B_{2n}\/}\nolimits\!\left(x\right)$ $\displaystyle=(-1)^{n+1}\frac{2(2n)!}{(2\pi)^{2n}}\sum_{k=1}^{\infty}\frac{% \mathop{\cos\/}\nolimits\!\left(2\pi kx\right)}{k^{2n}},$ 24.8.2 $\displaystyle\mathop{B_{2n+1}\/}\nolimits\!\left(x\right)$ $\displaystyle=(-1)^{n+1}\frac{2(2n+1)!}{(2\pi)^{2n+1}}\sum_{k=1}^{\infty}\frac% {\mathop{\sin\/}\nolimits\!\left(2\pi kx\right)}{k^{2n+1}}.$

The second expansion holds also for $n=0$ and $0.

If $n=1$ with $0, or $n=2,3,\dots$ with $0\leq x\leq 1$, then

 24.8.3 $\mathop{B_{n}\/}\nolimits\!\left(x\right)=-\frac{n!}{(2\pi i)^{n}}\sum_{% \begin{subarray}{c}k=-\infty\\ k\neq 0\end{subarray}}^{\infty}\frac{e^{2\pi ikx}}{k^{n}}.$

If $n=1,2,\dots$ and $0\leq x\leq 1$, then

 24.8.4 $\displaystyle\mathop{E_{2n}\/}\nolimits\!\left(x\right)$ $\displaystyle=(-1)^{n}\frac{4(2n)!}{\pi^{2n+1}}\sum_{k=0}^{\infty}\frac{% \mathop{\sin\/}\nolimits\!\left((2k+1)\pi x\right)}{(2k+1)^{2n+1}},$ 24.8.5 $\displaystyle\mathop{E_{2n-1}\/}\nolimits\!\left(x\right)$ $\displaystyle=(-1)^{n}\frac{4(2n-1)!}{\pi^{2n}}\sum_{k=0}^{\infty}\frac{% \mathop{\cos\/}\nolimits\!\left((2k+1)\pi x\right)}{(2k+1)^{2n}}.$

## §24.8(ii) Other Series

 24.8.6 $\displaystyle\mathop{B_{4n+2}\/}\nolimits$ $\displaystyle=(8n+4)\sum_{k=1}^{\infty}\frac{k^{4n+1}}{e^{2\pi k}-1},$ $n=1,2,\dots$, Symbols: $\mathop{B_{\NVar{n}}\/}\nolimits\!\left(\NVar{x}\right)$: Bernoulli polynomials, $e$: base of exponential function, $k$: integer and $n$: integer Referenced by: §24.8(ii) Permalink: http://dlmf.nist.gov/24.8.E6 Encodings: TeX, pMML, png See also: info for 24.8(ii) 24.8.7 $\displaystyle\mathop{B_{2n}\/}\nolimits$ $\displaystyle=\frac{(-1)^{n+1}4n}{2^{2n}-1}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{% e^{\pi k}+(-1)^{k+n}},$ $n=2,3,\dots$.

Let $\alpha\beta=\pi^{2}$. Then

 24.8.8 $\frac{\mathop{B_{2n}\/}\nolimits}{4n}\left(\alpha^{n}-(-\beta)^{n}\right)=% \alpha^{n}\sum_{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\alpha k}-1}-(-\beta)^{n}\sum% _{k=1}^{\infty}\frac{k^{2n-1}}{e^{2\beta k}-1},$ $n=2,3,\dots$.
 24.8.9 $\mathop{E_{2n}\/}\nolimits=(-1)^{n}\sum_{k=1}^{\infty}\frac{k^{2n}}{\mathop{% \cosh\/}\nolimits\!\left(\tfrac{1}{2}\pi k\right)}-4\sum_{k=0}^{\infty}\frac{(% -1)^{k}(2k+1)^{2n}}{e^{2\pi(2k+1)}-1},$ $n=1,2,\dots$.