Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$
Solution 1:
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.
Solution 2:
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
$$ L(\chi,2)=\sum_{k\geq 0}\left[\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right]=\frac{4\pi^2}{25\sqrt{5}}.$$
Solution 3:
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 $\sum \int = \int \sum$ .
Solution 4:
Note \begin{eqnarray} &&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)\\ &=&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2}\right) -\sum_{k=0}^\infty\left( \frac{1}{5k+3}-\frac{1}{5k+4} \right)\\ &=&\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}-\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}. \end{eqnarray} Let $$ f(x)=\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}x^{5k+2}, g(x)=\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}x^{5k+4}. $$ So \begin{eqnarray} f'(x)&=&\sum_{k=0}^\infty\frac{1}{5k+1}x^{5k+1}, f''(x)&=&\sum_{k=0}^\infty x^{5k}=\frac{1}{1-x^5}, \\ g'(x)&=&\sum_{k=0}^\infty\frac{1}{5k+3}x^{5k+3}, g''(x)&=&\sum_{k=0}^\infty x^{5k+2}=\frac{x^2}{1-x^5} \end{eqnarray} and hence \begin{eqnarray} &&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)\\ &=&\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}-\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}\\ &=&f(1)-g(1)\\ &=&\int_0^1\int_0^t\frac{1}{1-x^5}dxdt-\int_0^1\int_0^t\frac{x^2}{1-x^5}dxdt\\ &=&\int_0^1\int_x^1\frac{1}{1-x^5}dtdx-\int_0^1\int_x^1\frac{x^2}{1-x^5}dtdx\\ &=&\int_0^1\frac{1-x}{1-x^5}dx-\int_0^1\int_x^1\frac{(1-x)x^2}{1-x^5}dx\\ &=&\int_0^1\frac{1-x^2}{1+x+x^2+x^3+x^4}dx\\ &=&\int_0^1\frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+\frac{1}{x}+1+x+x^2}dx\\ &=&\int_0^1\frac{\frac{1}{x^2}-1}{(x+\frac{1}{x})^2+(x+\frac{1}{x})-1}dx\\ &=&-\int_0^1\frac{1}{(x+\frac{1}{x})^2+(x+\frac{1}{x})-1}d(x+\frac{1}{x})\\ &=&\int_2^\infty\frac{1}{u^2+u-1}du\\ &=&\frac{\log \left(\frac{1}{2} \left(7+3 \sqrt{5}\right)\right)}{2\sqrt{5}}. \end{eqnarray}