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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.

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  • $\begingroup$ Hint : calling $a_k$ the generic term of the series, compute $a_{k+1}+a_k$... $\endgroup$ – Tom-Tom Dec 11 '17 at 21:10
  • $\begingroup$ It reduces to $\sum_{n=1}^\infty \frac{e^{2i \pi nk/5}}{n} = -\log(1-e^{2i \pi k/5})$ $\endgroup$ – reuns Dec 12 '17 at 6:26
  • $\begingroup$ I find $\frac{2 \tanh ^{-1}\left(\frac{1}{\sqrt{5}}\right)}{\sqrt{5}}$ using the integral approach and simplifying the resulting log. $\endgroup$ – Dr. Wolfgang Hintze Dec 13 '17 at 11:53
  • $\begingroup$ Just an information: the Wikipedia article on the digamma function (en.wikipedia.org/wiki/Digamma_function) is very comprehensive, and in the paragraph "Series formula" is discusses the use of $\psi$ in calulating series which is very helpful in the present context (see my answer below). $\endgroup$ – Dr. Wolfgang Hintze Dec 14 '17 at 19:44
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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.

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  • $\begingroup$ Brilliant. I imagine you get something like $\int_0^1 1\frac{1-x^2}{1 + x + x^2 + x^3 + x^4}$ which you decompose into partial fraction by using the fact that $1-x^5$ has 5 roots on the unit circle? I wonder if there is a more geometric way to prove this. $\endgroup$ – gowrath Dec 11 '17 at 21:50
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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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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 $\sum \int = \int \sum$ .

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  • $\begingroup$ Why are able to do this? As in write $ \frac{1}{5k+j} $ as the integral? $\endgroup$ – bigfocalchord Dec 12 '17 at 0:25
  • $\begingroup$ @dydxx Because the indefinite integral of $x^{5k + j - 1}$ is equal to $1/(5k + j) x^{5k+j}$ and evaluating this from 0 to 1 gives $1/(5k + j)$. $\endgroup$ – gowrath Dec 12 '17 at 1:12
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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 $$ \begin{align} &\sum_{k=0}^\infty\left(\frac1{5k+1}-\frac1{5k+2}-\frac1{5k+3}+\frac1{5k+4}\right)\\ &=\scriptsize-\frac15\left(\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac45}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac35}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac25}\right)+\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac15}\right)\right)\\ &=-\frac15\left(H_{-4/5}-H_{-3/5}-H_{-2/5}+H_{-1/5}\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac2{\sqrt5}\log\left(\frac{1+\sqrt5}2\right)}\tag3 \end{align} $$

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  • $\begingroup$ I just found out why this seemed so familiar; it was because I'd done something very similar in this answer. $\endgroup$ – robjohn Dec 13 '17 at 20:16
  • $\begingroup$ Very nice idea to use telescoping series to produce harmonic numbers. At first sight I was surprised to see negative arguments in the harmonic numbers. I have derived the result from residues and found $\frac{5}{12}-\frac{1}{5} \left(H_{\frac{1}{5}}-H_{\frac{2}{5}}-H_{\frac{3}{5}}+H_{\frac{4}{5}}\right)$ which is equivalent. The $\frac{5}{12}$ is the summand with $k=0$ which is not covered by the residues of $H(-k)$. A question: how do you justify your last step? $\endgroup$ – Dr. Wolfgang Hintze Dec 14 '17 at 7:30
  • $\begingroup$ @Dr.WolfgangHintze: I used $(2)$ to evaluate each of $H_{-k/5}$ for $k\in\{1,2,3,4\}$. There is a lot of cancellation, and the end result is $(3)$. $\endgroup$ – robjohn Dec 14 '17 at 7:39
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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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  • $\begingroup$ I believe your limits should be $\int_2^\infty$ in the second last step? $\endgroup$ – gowrath Dec 19 '17 at 23:06
  • $\begingroup$ @gowrath, yes, you are right. Thanks. $\endgroup$ – xpaul Dec 20 '17 at 14:38
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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.

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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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Here is a generalization of this interesting problem.

Using residues we can find the sum for a given power $q$ of the fractions.

Let the sum to be calculated be

$$s(q)=f(q,0)+\sum _{k=1}^{\infty } f(q,k)$$

where the summand is

$$f(q,k)=(5 k+1)^{-q}-(5 k+2)^{-q}-(5 k+3)^{-q}+(5 k+4)^{-q}$$

We have extracted the summand with $k=0$ from the sum. The remaining sum

$$s_1(q) =\sum _{k=1}^{\infty } f(q,k) $$

can be calculated with the method explained in http://algo.inria.fr/flajolet/Publications/FlSa98.pdf from residues with the kernel function $H(-k)$ which essentially expresses an infinite sum by a finite sum over residues.

In our case it leads to

$$s_1(q))=-\sum _{j=1}^4 \text{res}\left(H_{-k} f(q,k),\left\{k,-\frac{j}{5}\right\}\right)$$

The notation means: take the residue of the expression $H_{-k} f(q,k)$ at the position $-\frac{j}{5}$ of the j-th pole of $f(q,k)$ in the complex k-plane and sum over all four poles $(j=1..4)$.

For the first two values of $q$ this gives for $s$

$$s(1) = \frac{5}{12}-\frac{\psi ^{(0)}\left(\frac{1}{5}\right)}{5}+\frac{\psi ^{(0)}\left(\frac{2}{5}\right)}{5}+\frac{\psi ^{(0)}\left(\frac{3}{5}\right)}{5}-\frac{\psi ^{(0)}\left(\frac{4}{5}\right)}{5}$$

$$s(2) = \frac{101}{144}+\frac{\psi ^{(1)}\left(\frac{1}{5}\right)}{25}-\frac{\psi ^{(1)}\left(\frac{2}{5}\right)}{25}-\frac{\psi ^{(1)}\left(\frac{3}{5}\right)}{25}+\frac{\psi ^{(1)}\left(\frac{4}{5}\right)}{25}$$

Typically, the polygamma function $\psi$ of degree $q-1$ and fractional argument. The generalization to higher values of $q$ is obvious.

Letting Mathematica do the simplification of the $\psi$-functions we obtain the following list in the format $\{q, s(q)\}$

$$ \begin{array}{l} \left\{1,\frac{\log \left(\frac{1}{2} \left(3+\sqrt{5}\right)\right)}{\sqrt{5}}\right\} \\ \left\{2,\frac{4 \pi ^2}{25 \sqrt{5}}\right\} \\ \left\{3,\frac{1475}{1728}-\frac{\psi ^{(2)}\left(\frac{1}{5}\right)}{250}+\frac{\psi ^{(2)}\left(\frac{2}{5}\right)}{250}+\frac{\psi ^{(2)}\left(\frac{3}{5}\right)}{250}-\frac{\psi ^{(2)}\left(\frac{4}{5}\right)}{250}\right\} \\ \left\{4,\frac{8 \pi ^4}{375 \sqrt{5}}\right\} \\ \left\{5,\frac{240275}{248832}-\frac{\psi ^{(4)}\left(\frac{1}{5}\right)}{75000}+\frac{\psi ^{(4)}\left(\frac{2}{5}\right)}{75000}+\frac{\psi ^{(4)}\left(\frac{3}{5}\right)}{75000}-\frac{\psi ^{(4)}\left(\frac{4}{5}\right)}{75000}\right\} \\ \end{array} $$

Better knowledge than mine about fractional argument polygamma function can possibly lead to further simplification.

The results for even $q$ have a particularly simple form:

$$ \begin{array}{l} \left\{2,\frac{4 \pi ^2}{25 \sqrt{5}}\right\} \\ \left\{4,\frac{8 \pi ^4}{375 \sqrt{5}}\right\} \\ \left\{6,\frac{536 \pi ^6}{234375 \sqrt{5}}\right\} \\ \left\{8,\frac{5776 \pi ^8}{24609375 \sqrt{5}}\right\} \\ \left\{10,\frac{3302008 \pi ^{10}}{138427734375 \sqrt{5}}\right\} \\ \end{array} $$

Notice that the result for $q=1$ (the original task) has been found by all contributors, that for $q=2$ was first provided here by Jack D'Aurizio with another method.

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