I'm reading Emil Artin's short book The Gamma Function.
Since page 29 he says,
Define $H(t) = \frac12 - t$ for $0<t<1$, $H(0)=0$ for $t=0$, and otherwise periodic of period 1. Then he introduces
$$H_{2n}(t):=2(-1)^{n-1}\sum_{i=1}^\infty\frac{\cos 2i\pi t}{(2i\pi)^{2n}}$$ $$H_{2n-1}(t):=2(-1)^{n}\sum_{i=1}^\infty\frac{\sin 2i\pi t}{(2i\pi)^{2n-1}}$$, then one can see $$-H_1(t)=H(t)$$ because $-H_1(t)$ is the Fourier series for $H(t)$ and $$H'_{n+1}(t)=H_n(t)$$ for $n\ge 2$ or $n=1$ for all non-integral $x$.
Till here all are good.
Then he says: if $n$ is even, $H_n(0)-H_n(t)$ has the same sign (plus or miuns) as $H_n(0)$ for all $t$.
I wonder how to prove this?
$$|H_{2n}(t)|\le 2\sum_{i=1}^{\infty}\frac{|\cos(2i\pi t)|}{(2i\pi)^{2n}}\le 2\sum_{i=1}^{\infty}\frac{1}{(2i\pi)^{2n}}=|H_{2n}(0)|$$
As $\forall t ||H_{2n}||_\infty\pm H_{2n}(t) \ge 0$ it follows the expected result i.e. $H_{2n}(0)-H_{2n}(t)$ has the same sign as $H_{2n}(0)$