<s>[INST] Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$$
and once this is done, to explain what this has to do with the construction of a pentagon (maybe some other polygon?) using a compass and straight edge.
In terms of evaluating the series, I tried writing each $n$ as $m \cdot 2^k$ and evaluating the summation there since $2^k$ will alternate between + and - mod 5. However, this leads to a divergent series and I think this is not a valid thing to do since the original series is not absolutely convergent so we can't rearrange terms like that.
 [/INST] 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
$$ L(\chi,2)=\sum_{k\geq 0}\left[\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right]=\frac{4\pi^2}{25\sqrt{5}}.$$
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