<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] 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}
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