Suppose $p$ and $q$ are positive integers, $p<q$.
How to prove this identity? $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\frac{\pi}{2}\csc\left(\frac{p\pi}{q}\right)+(-1)^{(p-1)}\sum_{n=1}^{(q-1)/2}2\cos\left(\frac{2\pi np}{q}\right)\ln\left(\cos\frac{n\pi}{q}\right)\quad \forall\;q \text{ is odd}\\ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\frac{\pi}{2}\csc\left(\frac{p\pi}{q}\right)+(-1)^{(p-1)}\sum_{n=1}^{q/2}2\cos\left(\frac{(2n-1)\pi p}{q}\right)\ln\left(\cos\frac{(2n-1)\pi}{2q}\right)\quad \forall\;q \text{ is even}$$ I have proved the first part for $q$ is odd by referring to the proof of Gauss's digamma theorm.
Using Abel’s limit theorem $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\lim_{t\rightarrow 1^-}\sum_{n=0}^{\infty}\frac{(-1)^n q}{p+qn}t^{p+qn}=(-1)^{-p}q\lim_{t\rightarrow 1^-}\sum_{n=0}^{\infty}\frac{(-t)^{p+qn}}{p+qn} $$ Using algorithm for extracting every $n^{th}$ term of a series, let $\omega=e^{2\pi i/q}$ $$ \sum_{n=0}^{\infty}\frac{(-t)^{p+qn}}{p+qn}=\frac{1}{q}\sum_{n=0}^{q-1}\omega^{-np}(-\ln(1+\omega^n t)) \\ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=(-1)^{-p}\lim_{t\rightarrow 1^-}\sum_{n=0}^{q-1}\omega^{-np}(-\ln(1+\omega^n t))=(-1)^{1-p}\sum_{n=0}^{q-1}\omega^{-np}\ln(1+\omega^n) $$ Replace $p$ by $q-p\quad$ ($q$ is odd) $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{q-p}{q}}=(-1)^{1-q+p}\sum_{n=0}^{q-1}\omega^{-n(q-p)}\ln(1+\omega^n)=(-1)^p\sum_{n=0}^{q-1}\omega^{np}\ln(1+\omega^n) $$
Subtract the two expressions to obtain $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1-\frac{p}{q}}=(-1)^{p-1}\sum_{n=0}^{q-1}2\cos\left(\frac{2\pi np}{q}\right)\ln(1+\omega^n) $$ Left side is real, so it is equal to the real part of right side. $$ \mathfrak{R}(\ln(1+\omega^n))=\mathfrak{R}\left(\ln\left(2\cos\frac{n\pi}{q}e^{i\frac{n\pi}{q}}\right)\right)=\ln\left | 2\cos\frac{n\pi}{q}\right | \\ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1-\frac{p}{q}}=(-1)^{p-1}\sum_{n=1}^{q-1}2\cos\left(\frac{2\pi np}{q}\right)\ln\left | \cos\frac{n\pi}{q}\right | $$ Since $$ \sum_{n=0}^{q-1}\cos\left(\frac{2\pi np}{q}\right)\ln(2)=0\quad \text{and}\quad \ln(\cos(0))=0 $$ Using the well-known identity $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1-\frac{p}{q}}=\pi\csc\left(\frac{p\pi}{q}\right)\ $$ Then $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\frac{\pi}{2}\csc\left(\frac{p\pi}{q}\right)+(-1)^{(p-1)}\sum_{n=1}^{q-1}\cos\left(\frac{2\pi np}{q}\right)\ln\left | \cos\frac{n\pi}{q}\right | \\ =\frac{\pi}{2}\csc\left(\frac{p\pi}{q}\right)+(-1)^{(p-1)}\sum_{n=1}^{(q-1)/2}2\cos\left(\frac{2\pi np}{q}\right)\ln\left(\cos\frac{n\pi}{q} \right)\quad\forall\;q \text{ is odd} $$ But I cannot apply the same proof process for $q$ is even. Can someone help me to complete the proof?
Thanks to Mr. metamorphy's answer. The proof is clever but it doesn't explain why my proof process for $q$ is odd can't be applied for $q$ is even. After a few days of trying, I manage to work it out for $q$ is even. The problem is when applying Abel’s limit theorem, different parity of $q$ comes out different result.
Using Abel’s limit theorem $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\lim_{t\rightarrow 1^-}\sum_{n=0}^{\infty}\frac{(-1)^n q}{p+qn}t^{p+qn}=(-1)^{-p}q\lim_{t\rightarrow 1^-}\sum_{n=0}^{\infty}\frac{(-1)^n(-t)^{p+qn}}{p+qn} \\ \sum_{n=0}^{\infty}\frac{(-1)^n(-t)^{p+qn}}{p+qn}=\sum_{n=0}^{\infty}\frac{(-t)^{p+2nq}}{p+2nq}-\sum_{n=0}^{\infty}\frac{(-t)^{p+(2n+1)q}}{p+(2n+1)q} $$ Using algorithm for extracting every $n^{th}$ term of a series, let $\omega=e^{2\pi i/2q}=e^{\pi i/q}$ $$ \sum_{n=0}^{\infty}\frac{q(-t)^{p+2nq}}{p+2nq}=\frac{q}{2q}\sum_{n=0}^{2q-1}\omega^{-np}(-\ln(1+\omega^n t))=-\frac{1}{2}\sum_{n=0}^{2q-1}\omega^{-np}\ln(1+\omega^n t) \\ \sum_{n=0}^{\infty}\frac{q(-t)^{p+(2n+1)q}}{p+(2n+1)q}=\sum_{n=0}^{\infty}\frac{q(-t)^{(p+q)+2nq}}{(p+q)+2nq}=-\frac{1}{2}\sum_{n=0}^{2q-1}\omega^{-n(p+q)}\ln(1+\omega^n t) \\ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\lim_{t\rightarrow 1^-}(-1)^{-p}\left ( \sum_{n=0}^{\infty}\frac{q(-t)^{p+2nq}}{p+2nq}-\sum_{n=0}^{\infty}\frac{q(-t)^{p+(2n+1)q}}{p+(2n+1)q} \right )\\ =\lim_{t\rightarrow 1^-}\frac{(-1)^{p-1}}{2}\left(\sum_{n=0}^{2q-1}\omega^{-np}\ln(1+\omega^nt)-\sum_{n=0}^{2q-1}\omega^{-n(p+q)}\ln(1+\omega^nt)\right) $$ For $n$ is odd, $\omega^{-n(q+p)}=-\omega^{-np}$ $$\omega^{-np}\ln(1+\omega^nt)-\omega^{-n(p+q)}\ln(1+\omega^nt)=2\omega^{-np}\ln(1+\omega^nt)$$ For $n$ is even, $\omega^{-n(q+p)}=\omega^{-np}$ $$\omega^{-np}\ln(1+\omega^nt)-\omega^{-n(p+q)}\ln(1+\omega^nt)=0$$ \begin{align} &\text{ }\quad\sum_{n=0}^{2q-1}\left ( \omega^{-np}\ln(1+\omega^nt)-\omega^{-n(p+q)}\ln(1+\omega^nt) \right )\\ &=\sum_{n=1,3,5\cdots ,2q-1}^{}2\omega^{-np}\ln(1+\omega^nt)= \sum_{n=1}^{q}2\omega^{-(2n-1)p}\ln(1+\omega^{(2n-1)}t)\\ \\ &\text{ }\quad\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\lim_{t\rightarrow 1^-}\frac{(-1)^{p-1}}{2}\sum_{n=1}^{q}2\omega^{-(2n-1)p}\ln(1+\omega^{(2n-1)}t)\\ &=(-1)^{p-1}\sum_{n=1}^{q}\omega^{-(2n-1)p}\ln(1+\omega^{(2n-1)}) \end{align} The rest of the proof is the same as the process for $q$ is odd, and goes to the desired result: $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+\frac{p}{q}}=\frac{\pi}{2}\csc\left(\frac{p\pi}{q}\right)+(-1)^{(p-1)}\sum_{n=1}^{q/2}2\cos\left(\frac{(2n-1)\pi p}{q}\right)\ln\left(\cos\frac{(2n-1)\pi}{2q}\right)\quad \forall\;q \text{ is even}$$
