Is there any similar solutions including $\pi$ like $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\cdots=\frac{\pi}{3\sqrt{3}}$? First equation is very popular - there are only odd numbers. Other words, numbers, which are  coprime with $2$.
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{7}+\cdots=\frac{\pi}{4}$$
Second is similar, but not very well known. There are numbers coprime with $3$.
$$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{8}+\cdots=\frac{\pi}{3\sqrt{3}}$$
With some simple transformation we can say, that
$$\left(1+\frac{1}{2}\right)\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{11}\right)\left(1-\frac{1}{13}\right)\cdots\right]^{-1}=\frac{\pi}{4}$$
where primes of form $4n+1$ is negative, other is positive. Similar way
$$\left(1+\frac{1}{3}\right)\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\right(1+\frac{1}{5}\left)\left(1-\frac{1}{7}\right)\left(1+\frac{1}{11}\right)\left(1-\frac{1}{13}\right)\cdots \right]^{-1}=\frac{\pi}{3\sqrt{3}}$$
where primes of form $6n+1$ is negative, other is positive. Here we can create a function $s_{m}(r)$, which generates $r$-th coprime with $m$, then
$$f_{k}(m)=\sum\limits_{i=0}^{\infty}\left[\sum\limits_{r=1}^{\varphi(m)/2}\left(\frac{1}{mi+s_{m}(2r-1)}\right)^k-\sum\limits_{r=1}^{\varphi(m)/2}\left(\frac{1}{mi+s_{m}(2r)}\right)^k\right]$$
$$g_{k}(m)=\prod\limits_{p}^{\infty}(1+p^{-k})^{-1}\cdot\prod\limits_{p|m}^{}(1+p^{-k})$$
$$h_{k}(mn+d)=\prod\limits_{p[mn+d]}^{\infty}\frac{p^k+1}{p^k-1}$$
$$j_{k}(m)=\prod\limits_{r=1}^{\varphi(m)/2}h_{k}(mn+s_{m}(2r-1))$$
$$f_{k}(m)=g_{k}(m)\cdot j_{k}(m)$$
Here $\varphi(m)$ - Euler totient function, $p[mn+d]$ - primes of form $mn+d$. Also
$$f_{k}(m)= f_{k}(m^2)= f_{k}(m^3)=\cdots= f_{k}(m^x)$$
so we can work only with $m$, which $|\mu(m)|=1$. If $m=2$, then we need to correct $mi$ in $f_{k}(m)$ to $2mi$ (and if $m=2$ or $m$ odd - $mn$ in $h_{k}(m)$ to $2mn$). Then we have
$$f_{1}(2)=\frac{\pi}{4}, f_{3}(2)=\frac{\pi^3}{32}, f_{5}(2)=\frac{5\pi^5}{1536}$$
Is there any results, including $\pi$, for example
$$f_{k}(5)=\pi^k\frac{a^{\frac{b}{c}}}{d^{\frac{e}{f}}}$$
where $a,b,c,d,e,f$ - integers?
If I made some mistakes, sorry for my English.
 A: Your first examples are $L$-series of quadratic Dirichlet characters.
If $\chi$ is a primitive Dirichlet character of conductor $N$
taking values in $\{\pm1\}$ and with $\chi(-1)=-1$ then
$$\zeta(s)L(s,\chi)=\zeta_K(s)$$
where $K=\Bbb Q(\sqrt{-N})$ and $\zeta_K$ is the Dedekind zeta
function of $K$. The analytic class number gives
$$L(1,\chi)=\frac{2\pi h}{w\sqrt{N}}$$
where $h$ is the classnumber of $K$ and $w$ the number of roots of unity in $K$.
For such $\chi$, one can obtain closed formulae for $L(m,\chi)$
whenever $m$ is a positive odd number. One way to so this is to use
the functional equation for the L-functions, to relate it to $L(1-m,\chi)$
which can be evaluated in terms of generalised Bernoulli numbers.
(See Washington's Introduction to Cyclotomic Fields).
A: I was going to write more or less the same answer of Lord Shark, so let us steer in a more elementary direction. The fact that Gregory series equals $\frac{\pi}{4}$ can be seen as a consequence of
$$ \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx = \int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}\tag{A}$$
but since
$$ \int_{0}^{+\infty}\frac{dx}{1+x^k} = \frac{\pi}{k\sin\frac{\pi}{k}}\tag{B}$$
from Euler's Beta function and the reflection formula for the $\Gamma$ function, a lot of similar identities can be simply derived from $(B)$ by reverse engineering. For instance, by taking $k=5$,
$$ \frac{\pi}{5\sin\frac{\pi}{5}}=\int_{0}^{1}\frac{1+x^3}{1+x^5}\,dx =\sum_{n\geq 0}(-1)^n\left(\frac{1}{5n+1}+\frac{1}{5n+4}\right)\tag{C}$$
and by taking $k=12$:
$$ \sum_{n\geq 0}(-1)^n\left(\frac{1}{12n+1}+\frac{1}{12n+11}\right)=\frac{\pi}{3\sqrt{2}(\sqrt{3}-1)}.\tag{D} $$
