Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ I'm looking for a way to prove
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$
I know that
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$
so maybe I could simplify the above more?

Because it wasn't explicitly stated, I probably should state that $E_{2m}$ are the Euler numbers.
Let's integrate the function $$ f(z) = \frac{\pi \csc (\pi z)}{(2z+1)^{2m+1}}, \quad m \in \mathbb{N}_{\ge 0},$$
around a square contour with vertices at $\pm (N+\frac{1}{2}) \pm i(N+\frac{1}{2})$ , where $N$ is a positive integer.
Letting $N$ go to infinity throught the positive integers, the integral vanishes (see here), and we end up with
$$ \begin{align} 0 &= \sum_{n=-\infty}^{\infty} \text{Res}[f(z),n] + \text{Res} \left[f(z),- \frac{1}{2} \right] \\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} + \text{Res}\left[f(z),-\frac{1}{2} \right]. \end{align}$$
Notice that
$$\text{Res} \left[ f(z),- \frac{1}{2} \right] = \text{Res} \left[f \left(z - \frac{1}{2}\right),0 \right] = \text{Res} \left[-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}},0 \right] .$$
Using the Maclaurin series of sec(z), that is,  $$ \sec (z) = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!}z^{2n} \, ,  \quad |x| < \frac{\pi}{2},$$
we have
$$-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}} = -\frac{\pi}{2^{2m+1}} \sum_{n=0}^{\infty} \frac{(-1)^{k} E_{2n}}{(2n)!} \pi^{2n} z^{2n-2m-1}  ,$$
from which we can conclude that
$$\text{Res} \left[-\frac{\pi \sec \pi z}{(2z)^{2m+1}},0 \right] = -\frac{\pi (-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m} .$$
Therefore,
$$2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} = \frac{(-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m+1}\, ,  $$
and the result follows.

EDIT:
Another approach is to use the partial fractions expansion
$$ \sec(z) = \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1) \pi}{\left(\frac{2n+1}{2} \right)^{2} \pi^{2}-z^{2}}. $$
For $|z| < \frac{\pi}{2}$, we have
$$ \begin{align}\sec(z) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1)\pi}{\left(\frac{2n+1}{2} \right)^{2}\pi^{2}} \frac{1}{1-\left( \frac{2z}{(2n+1)\pi}\right)^{2}} \\ &=\frac{4}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=0}^{\infty} \frac{(2z)^{2k}}{\left((2n+1) \pi\right)^{2k}}  \\ &= \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} . \end{align} $$
But we know that
$$ \sec(z) = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k} .$$
Therefore,
$$ \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k},$$
and the result follows by comparing the $k$th coefficient of both series.
