We have
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n}
\nonumber \\
& = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n} \int_0^\infty e^{-y} dy
\nonumber \\
& = \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} \int_0^\infty e^{-nx} dx
\nonumber \\
& = \frac{1}{2i} \sum_{n=1}^\infty \int_0^\infty (e^{-nx + \frac{i\pi n}{2}} - e^{-nx + \frac{-i\pi n}{2}}) dx
\nonumber \\
& = \frac{1}{2i} \int_0^\infty \left( \dfrac{1}{1-e^{-x + \frac{i\pi}{2}}} - \dfrac{1}{1-e^{-x + \frac{-i\pi}{2}}} \right) dx
\nonumber \\
& = \int_0^\infty \dfrac{1} {e^x + e^{-x}} dx
\nonumber \\
& = \frac{1}{2} \int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}
EDIT: Instead of using complex analysis to evaluate the integral, as I originally did, you can use the fact that ##\frac{d}{dx} \tan^{-1} (\sinh x) = \frac{1}{\cosh x}## to obtain:
\begin{align*}
\frac{1}{4} \int_{-\infty}^\infty \dfrac{2}{e^x + e^{-x}} dx = \frac{1}{4} [\tan^{-1} (\sinh x)]_{-\infty}^\infty = \frac{\pi}{4} .
\end{align*}
----
We will evaluate
\begin{align*}
\int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}
by considering the rectangular contour integral (see figure) of
\begin{align*}
\oint_C \dfrac{1} {e^z + e^{-z}} dz
\end{align*}
The integral along the vertical edges vanishes as:
\begin{align*}
f(z) = \dfrac{1}{e^{x+iy} + e^{-x-iy}} =
\begin{cases}
e^{- (x+iy)} & x \rightarrow \infty \\
e^{(x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}
So that
\begin{align*}
\oint_C \dfrac{1}{e^z + e^{-z}} dz & = \int_{-\infty}^\infty \dfrac{1}{e^x + e^{-x}} dx + \int_{-\infty+ i \pi}^{\infty + i \pi} \dfrac{1}{e^x + e^{-x}} dx
\nonumber \\
& = 2 \int_{-\infty}^\infty \dfrac{1} {e^x + e^{-x}} dx
\end{align*}
and so
\begin{align*}
\frac{1}{2} \int_{-\infty}^\infty \dfrac{1}{e^x + e^{-x}} dx & = \frac{i \pi}{2} \frac{1}{2 \pi i} \oint_C \dfrac{1}{e^z + e^{-z}} dz
\nonumber \\
& = \frac{i \pi}{2} \lim_{z \rightarrow \frac{i \pi}{2}} (z - \frac{i \pi}{2}) \dfrac{1}{e^z + e^{-z}}
\nonumber \\
& = \frac{i \pi}{2} \lim_{z \rightarrow \frac{i \pi}{2}} \dfrac{1}{e^z - e^{-z}}
\nonumber \\
& = \frac{\pi}{4} .
\end{align*}
So finally,
\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} = \frac{\pi}{4} .
\end{align*}