Suppose we seek to show that
$$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}
\frac{1}{\exp((2k+1)\pi/2)+\exp(-(2k+1)\pi/2)}
= \frac{\pi}{16}.$$
The sum term
$$S(x) = \sum_{k\ge 1} \frac{(-1)^{k+1}}{2k-1}
\frac{1}{\exp(x(2k-1))+\exp(-x(2k-1))}$$
is harmonic and may be evaluated by inverting its Mellin transform.
We are interested in $S(\pi/2).$
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)= \mathfrak{M}(g(x);s)$
is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{(-1)^{k+1}}{2k-1},
\quad \mu_k = 2k-1 \quad \text{and} \quad
g(x) = \frac{1}{\exp(x)+\exp(-x)}.$$
We need the Mellin transform $g^*(s)$
of $g(x)$ which is computed as follows:
$$g^*(s) = \int_0^\infty \frac{1}{\exp(x)+\exp(-x)} x^{s-1} dx
= \int_0^\infty \frac{\exp(-x)}{1+\exp(-2x)} x^{s-1} dx
\\ = \int_0^\infty \sum_{q\ge 0} (-1)^q \exp(-(2q+1)x) x^{s-1} dx
= \Gamma(s) \sum_{q\ge 0} \frac{(-1)^q}{(2q+1)^s}
= \Gamma(s) \beta(s)$$
where $$\beta(s) = 4^{-s}
\left(\zeta\left(s,\frac{1}{4}\right)
-\zeta\left(s,\frac{3}{4}\right)\right).$$
Note that $\beta(s)$ does not have a pole at $s=1$ and
$\beta(1) = \frac{\pi}{4}.$
Hence the Mellin transform $Q(s)$ of $S(x)$ is given by
$$ Q(s) = \Gamma(s) \beta(s) \beta(s+1)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \beta(s+1)$$
where $\Re(s) > 0$.
Intersecting the fundamental strip and the half-plane from the zeta
function term we find that the Mellin inversion integral for an
expansion about zero is
$$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds$$
which we evaluate in the left half-plane $\Re(s)<1/2.$
The two beta function terms cancel the poles of the gamma function
term and we are left with just
$$\mathrm{Res}(Q(s)/x^s; s=0) = \frac{\pi}{8}.$$
This shows that
$$S(x) = \frac{\pi}{8}
+ \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds.$$
To treat the integral recall the duplication formula of the gamma
function:
$$\Gamma(s) =
\frac{1}{\sqrt\pi} 2^{s-1}
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right).$$
which yields for $Q(s)$
$$\frac{1}{\sqrt\pi} 2^{s-1}
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right)
\beta(s) \beta(s+1)$$
Furthermore observe the following variant of the functional equation
of the Riemann zeta function adapted to the beta function:
$$\Gamma\left(\frac{s+1}{2}\right)\beta(s)
= 2^{1-2s} \pi^{s-1/2} \Gamma\left(1-\frac{s}{2}\right)
\beta(1-s)$$
which gives for $Q(s)$
$$\frac{1}{\sqrt\pi} 2^{s-1}
\Gamma\left(\frac{s}{2}\right)
2^{1-2s} \pi^{s-1/2} \Gamma\left(1-\frac{s}{2}\right)
\beta(1-s)\beta(s+1)
\\ = 2^{-s} \pi^{s-1} \frac{\pi}{\sin(\pi s/2)}
\beta(1-s)\beta(s+1)
\\ = 2^{-s} \frac{\pi^s}{\sin(\pi s/2)}
\beta(1-s)\beta(s+1).$$
Now put $s=-u$ in the remainder integral to get
$$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty}
2^u \frac{\pi^{-u}}{\sin(-\pi u/2)}
\zeta(1+u)\zeta(1-u) x^u du
\\ = -\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty}
2^{-u} \frac{\pi^{u}}{\sin(\pi u/2)}
\zeta(1+u)\zeta(1-u) (4 x / \pi^2)^u du.$$
We have shown that
$$S(x) = \frac{\pi}{8} - S(\pi^2/4/x).$$
In particular we get
$$S(\pi/2) = \frac{\pi}{8} - S(\pi/2)
\quad\text{or}\quad
S(\pi/2) = \frac{\pi}{16}$$
as claimed.
Addendum. The functional equation for $\beta(s)$ can be derived
from the functional equation of the Hurwitz Zeta function:
$$\zeta\left(1-s, \frac{m}{n}\right)
= \frac{2\Gamma(s)}{(2\pi n)^s}
\sum_{k=1}^n
\left[\cos\left(\frac{\pi s}{2} - \frac{2\pi k m}{n}\right)
\zeta\left(s,\frac{k}{n}\right)\right].$$
where $1\le m\le n.$
This yields
$$\zeta(1-s, 1/4) =
\frac{2\Gamma(s)}{2^{3s}\pi^s}
\left[ \cos(\pi s/2 - \pi/2) \zeta(s, 1/4)
+ \cos(\pi s/2 - \pi) \zeta(s, 1/2)
\\ + \cos(\pi s/2 - 3\pi/2) \zeta(s, 3/4)
+ \cos(\pi s/2 - 2\pi) \zeta(s)\right]
\\ = \frac{2\Gamma(s)}{2^{3s}\pi^s}
\left[ \sin(\pi s/2) \zeta(s, 1/4) - \cos(\pi s/2) \zeta(s, 1/2)
\\ -\sin(\pi s/2) \zeta(s, 3/4) + \cos(\pi s/2) \zeta(s)\right].$$
Similarly
$$\zeta(1-s, 3/4) =
\frac{2\Gamma(s)}{2^{3s}\pi^s}
\left[ \cos(\pi s/2 - 3\pi/2) \zeta(s, 1/4)
+ \cos(\pi s/2 - 3\pi) \zeta(s, 1/2)
\\ + \cos(\pi s/2 - 9\pi/2) \zeta(s, 3/4)
+ \cos(\pi s/2 - 6\pi) \zeta(s)\right]
\\ = \frac{2\Gamma(s)}{2^{3s}\pi^s}
\left[- \sin(\pi s/2) \zeta(s, 1/4) - \cos(\pi s/2) \zeta(s, 1/2)
\\ +\sin(\pi s/2) \zeta(s, 3/4) + \cos(\pi s/2) \zeta(s)\right].$$
Subtract to obtain
$$4^{1-s} \beta(1-s)
= \frac{2\Gamma(s)}{2^{3s}\pi^s}
(2\sin(\pi s/2) \zeta(s, 1/4) - 2\sin(\pi s/2) \zeta(s, 3/4))$$
or
$$4^{1-s} \beta(1-s) = \frac{2\Gamma(s)}{2^{3s}\pi^s} 2\sin(\pi s/2)
4^s \beta(s)$$
which is
$$\beta(1-s) =
2^s \frac{\Gamma(s)}{\pi^s} \sin(\pi s/2) \beta(s)
\\ = \frac{1}{\pi^{s+1/2}} 2^{2s-1}
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right)
\sin(\pi s/2) \beta(s)
\\ = \frac{1}{\pi^{s+1/2}} 2^{2s-1}
\frac{\pi}{\sin(\pi s/2)} \Gamma\left(1-\frac{s}{2}\right)^{-1}
\Gamma\left(\frac{s+1}{2}\right)
\sin(\pi s/2) \beta(s)$$
which yields
$$\beta(1-s)\Gamma\left(1-\frac{s}{2}\right)
= \pi^{1/2-s} 2^{2s-1}
\beta(s) \Gamma\left(\frac{s+1}{2}\right)$$
which is the desired result.
