<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] Not an answer but too long for a comment.
Considering
$$f(i)=\sum_{k=0}^p \frac 1 {5k+i}=\frac{1}{5} \left(\psi ^{(0)}\left(\frac{i}{5}+p+1\right)-\psi
   ^{(0)}\left(\frac{i}{5}\right)\right)$$ then $5\left(f(1)-f(2)-f(3)+f(4)\right)$ write$$\psi ^{(0)}\left(p+\frac{6}{5}\right)-\psi ^{(0)}\left(p+\frac{7}{5}\right)-\psi
   ^{(0)}\left(p+\frac{8}{5}\right)+\psi ^{(0)}\left(p+\frac{9}{5}\right)-\psi
   ^{(0)}\left(\frac{4}{5}\right)+\psi ^{(0)}\left(\frac{3}{5}\right)+\psi
   ^{(0)}\left(\frac{2}{5}\right)-\psi ^{(0)}\left(\frac{1}{5}\right)$$ which, using  generalized harmonic numbers, reduces to 
$$5\left(f(1)-f(2)-f(3)+f(4)\right)=H_{p+\frac{1}{5}}-H_{p+\frac{2}{5}}-H_{p+\frac{3}{5}}+H_{p+\frac{4}{5}}+2 \sqrt{5}
   \coth ^{-1}\left(\sqrt{5}\right)$$ Now, using the asymptotics of generalized harmonic numbers, we end with 
$$\sum_{k=0}^p\left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)=\frac{2 \coth ^{-1}\left(\sqrt{5}\right)}{\sqrt{5}}-\frac{2}{125 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit.
It also allows to compute partial sums. Using $p=10$, the exact value is $\frac{2826908374432441763845}{6569974349001513017568}\approx 0.430277$ while the above development gives $\approx 0.430249$. 
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