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

Another approach is to use the Anti-difference concept.
Since the digamma function is defined as
$$
\psi \left( z \right) = {d \over {dz}}\ln \Gamma \left( z \right)
$$
its functional equation is
$$
\Delta \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {d \over {dz}}\ln \left( {z\,\Gamma \left( z \right)} \right) - {d \over {dz}}\ln \left( {\Gamma \left( z \right)} \right)
 = {d \over {dz}}\ln z = {1 \over z}
$$
It follows that
$$
\sum\limits_{n = 1}^N {{1 \over {n + a}}}  = \sum\limits_{n = 1}^N {\psi \left( {n + a + 1} \right) - \psi \left( {n + a} \right)}  = \psi \left( {N + a + 1} \right) - \psi \left( {1 + a} \right)
$$
and therefore
$$
\sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}}  = {1 \over 4}\sum\limits_{n = 1}^N {{1 \over {n + 1/4}} - {1 \over n}}
  = {1 \over 4}\left( {\psi \left( {N + 5/4} \right) - \psi \left( {5/4} \right) - \psi \left( {N + 1} \right) + \psi \left( 1 \right)} \right)
$$
Since $\psi(z)$ is holomorphic for $0<\Re(z)$, then
$$
\psi \left( {N + 5/4} \right) - \psi \left( {N + 1} \right) = \psi ^{\,\left( 1 \right)} \left( {N + 1} \right){1 \over 4} + {{\psi ^{\,\left( 2 \right)} \left( {N + 1} \right)} \over {2!}}\left( {{1 \over 4}} \right)^2  +  \cdots 
$$
and since
$$
\mathop {\lim }\limits_{N \to \infty } \psi ^{\,\left( k \right)} \left( {N + 1} \right) = 0\quad \left| {\;1 \le k} \right.
$$
Therefore
$$
\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}}  = {1 \over 4}\left( {\psi \left( 1 \right) - \psi \left( {5/4} \right)} \right) = {3 \over 4}\ln 2 + {\pi  \over 8} - 1
$$
