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.
 A: 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.
A: 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
$$ L(\chi,2)=\sum_{k\geq 0}\left[\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right]=\frac{4\pi^2}{25\sqrt{5}}.$$
A: 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 $\sum \int = \int \sum$ .
A: 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}
A: Preliminaries
Using the identity
$$
\frac1n\sum_{k=0}^{n-1} e^{2\pi ik(j-m)/n}=[j\equiv m\pmod{n}]\tag1
$$
we get the following formula for Extended Harmonic Numbers with rational indices.
Let $0\le m\lt n$, then
$$
\begin{align}
H_{-m/n}
&=\sum_{j=1}^\infty\left(\frac1j-\frac1{j-m/n}\right)\\
&=n\sum_{j=1}^\infty\left(\frac1{jn}-\frac1{jn-m}\right)\\
&=\sum_{k=0}^{n-1}\sum_{j=1}^\infty\left(e^{2\pi ijk/n}-e^{2\pi ik(j+m)/n}\right)\frac1j\\
&=\sum_{k=1}^{n-1}\left(1-e^{2\pi ikm/n}\right)\sum_{j=1}^\infty e^{2\pi ijk/n}\frac1j\\
&=-\sum_{k=1}^{n-1}\left(1-e^{2\pi ikm/n}\right)\log\left(1-e^{2\pi ik/n}\right)\\
&=\sum_{k=1}^{n-1}e^{\pi ikm/n}2i\sin\left(\frac{\pi km}n\right)\left(\frac{\pi ik}n+\log\left(-2i\sin\left(\frac{\pi k}n\right)\right)\right)\\
&=\sum_{k=1}^{n-1}e^{\pi i(km/n+1/2)}2\sin\left(\frac{\pi km}n\right)\left(\pi i\left(\frac kn-\frac12\right)+\log\left(2\sin\left(\frac{\pi k}n\right)\right)\right)\\
&=\bbox[5px,border:2px solid #C0A000]{-\sum_{k=1}^{n-1}\left[\left(1-\cos\left(\frac{2\pi km}n\right)\right)\log\left(2\sin\left(\frac{\pi k}n\right)\right)+\sin\left(\frac{2\pi km}n\right)\left(\frac kn-\frac12\right)\pi\right]}\tag2
\end{align}
$$
Mathematica implementation of $(2)$:
h[m_,n_]:=-Sum[(1-Cos[2Pi k m/n])Log[2Sin[Pi k/n]]+Sin[2Pi k m/n](k/n-1/2)Pi,{k,1,n-1}]

Application to the Question
$$
\begin{align}
&\sum_{k=0}^\infty\left(\frac1{5k+1}-\frac1{5k+2}-\frac1{5k+3}+\frac1{5k+4}\right)\\
&=\scriptsize-\frac15\left(\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac45}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac35}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac25}\right)+\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac15}\right)\right)\\
&=-\frac15\left(H_{-4/5}-H_{-3/5}-H_{-2/5}+H_{-1/5}\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac2{\sqrt5}\log\left(\frac{1+\sqrt5}2\right)}\tag3
\end{align}
$$
A: This is more a comment.
These techniques work for
any sum of the form
$\sum_{i=0}^{\infty} \sum_{j=1}^{m} \dfrac{a_j}{mi+j}.
$
This case is
$m=5, a_j = 1, -1, -1,1, 0$.
These sums converge
if and only if
$\sum_{j=1}^m a_j = 0
$.
The convergence criteria
can be proved in
an elementary way.
To get a formula
for the sum,
one way is to use
multisection of series.
Another is the
integration formula used
in some of the answers.
A: 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 $\approx 0.430249$. 
