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.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$

If we call our sum $S$, then we see that
\begin{align*}
S&=\sum_{n=1}^{\infty}\left(\frac{1}{4n+1}-\frac{1}{4n}\right)\\
&=-\frac{1}{4}\left(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+\frac{1}{4}}\right)-\gamma+\gamma\right)
\end{align*}
where $\gamma$ is the Euler-Mascheroni constant. The reason that I have done this reformulation of the problem is that there is a well known function called the digamma function defined by
$$\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$$
where $\Gamma(s)$ is the gamma function. It is well known that
$$\psi(s+1)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+s}\right)-\gamma$$
and so we get now that
$$S=-\frac{1}{4}\left(\psi(5/4)+\gamma\right)$$
Gauss's formula relates values of the digamma function at rational numbers to a finite number of elementary functions, namely that
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln(2m)-\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right)+2\sum_{n=1}^{\lfloor (m-1)/2\rfloor}\cos\left(\frac{2\pi nr}{m}\right)\ln\left(\sin\left(\frac{\pi n}{m}\right)\right)$$
This formula is only valid for $r<m$, and so by using the reccurence relation
$$\psi(s+1)=\psi(s)+\frac{1}{s}$$
we get that
\begin{align*}
\psi\left(\frac{5}{4}\right)&=\psi\left(\frac{1}{4}\right)+4\\
&=-\gamma-\ln\left(2\left(4\right)\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\sum_{n=1}^{\lfloor(3)/2\rfloor}\cos\left(\frac{2\pi n}{4}\right)\ln\left(\sin\left(\frac{\pi n}{4}\right)\right)+4\\
&=-\gamma-\ln\left(8\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\cos\left(\frac{\pi}{2}\right)\ln\left(\sin\left(\frac{\pi}{4}\right)\right)+4
\end{align*}
substituting for our basic trig values and using properties of natural logs this simplifies to
$$\psi\left(\frac{5}{4}\right)=-\gamma-3\ln\left(2\right)-\frac{\pi}{2}+4$$
Plugging this back into our main formula for $S$ in terms of the digamma function yields that
$$S=\frac{1}{8}\left(6\ln\left(2\right)+\pi-8\right)$$
which completes our calculation. In general, this method can be used to solve any sum in form
$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+c}\right)$$
where $c$ is rational.
