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

This series can be evaluated with some generating function techniques. It will be related to the well-known generating function
$$ H(x) = \sum_{n=1}^\infty \frac{1}{n} x^n $$
which has closed form found by:
$$ H'(x) = \sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$
$$ H(x) = \int\! \frac{dx}{1-x} = \ln(1-x) + C = \ln(1-x) $$
since $H(0)=0$ requires $C=0$. For this problem, we'll want $\ln$ to represent the principal branch of the complex function.
But the function $f$ involves only subsequences of $1/n$, not every term. A standard technique for dealing with subsequences in generating functions is a selector function for roots of unity:
$$ \sigma_k(x) = \frac{1}{k} \sum_{\ell=0}^{k-1} x^{\ell} $$
which has the property that for every $m \in \mathbb{Z}$,
$$ \sigma_k \! \left(e^{2 \pi m i/k}\right) = \begin{cases}
1 & \quad\mathrm{if}\ m \equiv 0 \pmod{k} \\
0 & \quad\mathrm{if}\ m \not\equiv 0 \pmod{k} \end{cases} $$
To match the series terms, note that
$$ \forall m\in \mathbb{Z}: \sigma_4(i^{m-1}) - \sigma_4(i^m) =
\begin{cases} 1 & \quad\mathrm{if}\ m \equiv 1 \pmod{4} \\
-1 & \quad\mathrm{if}\ m \equiv 0 \pmod{4} \\
0 & \quad\mathrm{otherwise} \end{cases} $$
So define
$$ f(x) = \sum_{m=4}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$
and the desired infinite sum will be $f(i)$.
$$ f(x) = -1 + \sum_{m=1}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$
$$ f(x) = -1 + \sum_{m=1}^\infty \frac{1}{4m} (1 + x^{m-1} + x^{2m-2} + x^{3m-3} - 1 - x^m - x^{2m} - x^{3m}) $$
$$ f(x) = -1 + \frac{1}{4} \sum_{m=1}^\infty \left[ (x^{-1} - 1)\frac{x^m}{m} +
(x^{-2} - 1)\frac{x^{2m}}{m} + (x^{-3} - 1)\frac{x^{3m}}{m} \right] $$
$$ f(x) = -1 + \frac{1}{4}(x^{-1} - 1)\ln(1-x) +
 \frac{1}{4}(x^{-2} - 1)\ln(1-x^2) +
 \frac{1}{4}(x^{-3} - 1)\ln(1-x^3) $$
Then the infinite sum is
$$ f(i) = -1 + \frac{-1-i}{4} \ln(1-i) - \frac{2}{4} \ln 2 + \frac{-1+i}{4}\ln(1+i) $$
Since $1-i = \sqrt{2}\,e^{-\pi i/4}$ and $1+i = \sqrt{2}\,e^{\pi i/4}$,
$$ f(i) = -1 + \frac{-1-i}{4}\left(\frac{1}{2} \ln 2 - \frac{\pi i}{4}\right)
 - 2 \ln 2 + \frac{-1+i}{4}\left(\frac{1}{2} \ln 2 + \frac{\pi i}{4}\right) $$
Multiplying out and collecting like terms cancels all the imaginary parts. (If they didn't cancel, we'd know something was wrong with the calculations, since the original sum is clearly real!) So finally, the sum is
$$ f(i) = -1 + \frac{\pi}{8} - \frac{3}{4} \ln 2 $$
