I am looking at the following result here on page 121.
The Fourier series of the $p$'th Periodic Bernoulli polynomial $\mathcal{P}_p(x) := \mathcal{B}_p(\{x\})$ ($p$'th Bernoulli polynomial evaluated at the fractional part $\{x\} \equiv x-\lfloor x\rfloor$) is given as:
$$\frac{\mathcal{P}_p(x)}{p!} = -2 \sum_{k=1}^{\infty}\frac{\cos(2k \pi x -p \frac{\pi}{2}) }{(2k\pi)^p}$$
The author then derives this estimate for Bernoulli numbers from the Fourier series:
$$2 \frac{(2j)!}{(2\pi)^{2j}} < B_j < 4 \frac{(2j)!}{(2\pi)^{2j}}$$
and $\|\mathcal{P}_p(x)\|_{\infty} \equiv \max\limits_{x \in [0,1]}|\mathcal{P}_p(x)| < 4(p!)(2\pi)^{-p} $
How is this done? I wanted to try the recovery formulas but the series isn't of the form $\frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_k \sin(kx)$. Even assuming the second I am not sure how to get the estimate for the maximum.
Any help appreciated.
Shifting the argument of $\cos$ by $p \frac{\pi}{2}$ just gives a sign, and makes it a $\sin$ if $p$ is odd. If $p = 2m$ then
$$\cos \bigl(\alpha - p\tfrac{\pi}{2}\bigr) = \cos (\alpha - m\pi) = (-1)^m\cos \alpha\,,$$
and if $p = 2m+1$ then
$$\cos \bigl(\alpha - p\tfrac{\pi}{2}\bigr) = \cos \bigl(\alpha - m\pi - \tfrac{\pi}{2}\bigr) = (-1)^m\cos \bigl(\alpha - \tfrac{\pi}{2}\bigr) = (-1)^m\sin \alpha\,.$$
Using the shift allows one to express the Fourier series in a uniform manner for odd and even $p$. I prefer using the complex form of the Fourier series, where this problem doesn't arise. With
\begin{align} 2\cos \bigl(\alpha - p\tfrac{\pi}{2}\bigr) &= \exp \bigl(i\alpha - ip\tfrac{\pi}{2}\bigr) + \exp \bigl(-i\alpha + ip\tfrac{\pi}{2}\bigr) \\ &= (-i)^p e^{i\alpha} + i^p e^{-i\alpha} \end{align}
we find
\begin{align} -2\frac{\cos \bigl(2k\pi x - p\frac{\pi}{2}\bigr)}{(2k\pi)^p} &= -\frac{(-i)^p e^{2\pi i kx}}{(2k\pi)^p} - \frac{i^p e^{-2\pi ikx}}{(2\pi k)^p} \\ &= -\frac{e^{2\pi ikx}}{(2\pi ik)^p} - \frac{e^{2\pi i(-k)x}}{(2\pi i(-k))^p} \end{align}
and thus the complex form of the Fourier series
$$\frac{\mathcal{P}_p(x)}{p!} = - \sum_{k \neq 0} \frac{e^{2\pi ikx}}{(2\pi ik)^p}\,.$$
From this we obtain a bound for $\lvert \mathcal{P}_p(x)\rvert$ by taking the absolute value in each term of the sum when $p > 1$. For even $p$, since then $(-k)^p = k^p$, this gives the exact value of $\lvert \mathcal{P}_p(0)\rvert$ and thus a sharp bound, while for odd $p$ the obtained bound is not sharp, but it overestimates $\sup \:\{ \lvert \mathcal{P}_p(x)\rvert : x \in [0,1]\}$ not too badly. The bound we obtain is
$$\lvert \mathcal{P}_p(x)\rvert \leqslant \frac{p!}{(2\pi)^p}\sum_{k \neq 0} \frac{1}{\lvert k\rvert^p} = 2\frac{p!}{(2\pi)^p}\zeta(p)\,.$$
Since $1 < \zeta(p) \leqslant \zeta(2) = \frac{\pi^2}{6} < 2$ for $p \geqslant 2$ we thus have the upper bound
$$\lvert \mathcal{P}_p(x)\rvert \leqslant \frac{\pi^2}{3}\cdot \frac{p!}{(2\pi)^p} < 4\frac{p!}{(2\pi)^p}$$
for all $p > 1$ (one verifies that this bound also holds for $p = 1$), and the lower bound
$$\lvert \mathcal{P}_p(0)\rvert > 2\frac{p!}{(2\pi)^p}$$
for even $p$.