Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$

This is $L(1,\chi)$ where $\chi$ is the quadratic Dirichlet character of conductor $5$ defined by $\chi(a)=\left(\frac a5\right)$. Texts on number theory such as Washington's Introduction to Cyclotomic Fields will give details on how to evaluate these.

For a more naive approach, note that your sum is $$\sum_{n=0}^\infty\int_0^1(x^{5n}-x^{5n+1}-x^{5n+2}+x^{5n+3})\,dx =\int_0^1\frac{1-x-x^2+x^3}{1-x^5}\,dx.$$ You can use your favourite integration methods to tackle this.

As a followup to Lord Shark's answer, the idea of using $\frac{1}{n+1}=\int_{0}^{1}x^n\,dx $ can be naive but it is pretty effective. Once we have $$ L(\chi,1)=\sum_{n\geq 1}\frac{\left(\frac{n}{5}\right)}{n}=\int_{0}^{1}\frac{1-x^2}{1+x+x^2+x^3+x^4}\,dx $$ the integral can be evaluated by partial fraction decomposition, since $\int_{0}^{1}\frac{dx}{x-\xi}=\log\left(1-\frac{1}{\xi}\right) $.
Let $\omega=\exp\left(\frac{2\pi i}{5}\right)$. We have $$\begin{eqnarray*} \operatorname*{Res}_{x=\omega^k}\frac{1-x^2}{1+x+x^2+x^3+x^4}&=&\lim_{x\to \omega^k}\frac{(1-x-x^2+x^3)(x-\omega^k)}{1-x^5}\\&\stackrel{d.H.}{=}&\lim_{x\to \omega^k}\frac{-1-2x+3x^2}{-5x^4}\\&=&\frac{1}{5}\lim_{x\to \omega^k}\left(x+2x^2-3x^3\right)\end{eqnarray*} $$ for any $k\in[1,4]$, hence $$\begin{eqnarray*} L(\chi,1) &=& \frac{1}{5}\sum_{k=1}^{4}\left(\omega^k+2\omega^{2k}-3\omega^{3k}\right)\log(1-\omega^{-k})\\&=&\color{red}{\frac{2\log(5+\sqrt{5})-\log(20)}{\sqrt{5}}}.\end{eqnarray*}$$ In a similar way you may prove that $$ L(\chi,2)=\sum_{k\geq 0}\left[\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right]=\frac{4\pi^2}{25\sqrt{5}}.$$

I've done this before. Write: $\dfrac{1}{5k+j} = \displaystyle \int_{0}^1 x^{5k+j-1}dx, j = 2,3,4,5$, and compute the sum of integrand, and can use some powerful DCT theorem, and also $\sum \int = \int \sum$ .