How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$

May be interesting to consider the most general case of
$$S_p=\sum _{n=1}^{p } \left(\frac{1}{a n+b}-\frac{1}{c n+d}\right)$$ Using the digamma function,
$$S_p=\frac{c \psi \left(\frac{b}{a}+p+1\right)-a \psi \left(\frac{d}{c}+p+1\right)-c \psi
   \left(\frac{b}{a}+1\right)+a
   \psi\left(\frac{d}{c}+1\right)}{a c}$$ Expanded as series for large values of $p$
$$S_p=\frac{(c-a) \log (p)-c \psi \left(\frac{a+b}{a}\right)+a \psi
   \left(\frac{c+d}{c}\right)}{a
   c}+O\left(\frac{1}{p}\right)$$ which can converge only if $c=a$. In such a case
$$S_p=\frac{\psi \left(\frac{a+d}{a}\right)-\psi
   \left(\frac{a+b}{a}\right)}{a}+O\left(\frac{1}{p}\right)$$
For the particular case where $d=0$, $\psi(1)=-\gamma$  and
$$S_p=-\frac 1a\left(\psi \left(1+\frac{b}{a}\right)+\gamma \right)+O\left(\frac{1}{p}\right)$$ and if $a=k b$ the term in parentheses has simple expressions up to ... $k=4$.
