I would like to show that:

$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$

We have:

$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$

I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...

  • 1
    $\begingroup$ Technically, $\arctan x$ can be written as a power series valid around any given point. It's just that the Maclaurin series - the one with the coefficients you want - only converges in $[-1,1]$. Also, it happens that your series is $L(1,\chi_2)$, where $\chi_2$ is the unique nontrivial character modulo 3. $\endgroup$ – anon Feb 22 '12 at 21:20

Regularized the series: $$ \begin{eqnarray} \sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} &=& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) = \sum_{n=0}^m \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\ &=& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x = \int_0^1 \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x \end{eqnarray} $$ Now we can take the limit by dominating convergence theorem: $$ \sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2 \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}} $$

  • $\begingroup$ Clever. The actual computations are probably not so terribly different from my idea. $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:26
  • $\begingroup$ Proceeding per your suggestion leads to more complicated integrals: $\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \mathrm{d} x \int_0^x \mathrm{d} y \sum_{n=0}^\infty y^{3n} = \int_0^1 \mathrm{d} x \int_0^x \frac{\mathrm{d} y }{1-y^3}$. $\endgroup$ – Sasha Feb 22 '12 at 21:42
  • $\begingroup$ Um, yeah. The inner one leads to one integral like yours plus a logarithm, but the outer one will then take more work. You win. $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:46
  • $\begingroup$ Thank you for this nice answer! However woundn't it rather be: $$ \int_0^1 \frac{1-x^{3(m+1)}}{1+x + x^2} \mathrm{d} x $$ ? $\endgroup$ – Chon Feb 23 '12 at 7:53
  • $\begingroup$ Yes, you are correct. I will edit the post. $\endgroup$ – Sasha Feb 23 '12 at 13:31

What do you get if you differentiate $$\sum_{n=0}^\infty \frac{x^{3n+2}}{(3n+1)(3n+2)}$$ twice?

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    $\begingroup$ Why do I like answering a question with a question? $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:23
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    $\begingroup$ Why not? [And now the system makes me spoil the joke by forcing me to type more.] $\endgroup$ – André Nicolas Feb 22 '12 at 21:36
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    $\begingroup$ Too bad about your spoiled joke. But to be serious, I think it is pedagogically sound. [I tried responding “Why?” ending with a bunch of non-breakable spaces, but the system saw through the subterfuge and rejected it.] $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:43
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    $\begingroup$ Type \ \ \ \ \ \ between dollar signs. $\endgroup$ – David Mitra Feb 22 '12 at 21:56
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    $\begingroup$ @AndréNicolas Use \color{white}{\text{ and some extra space here }} and no one will notice. $\endgroup$ – Simply Beautiful Art Oct 12 '16 at 21:00

An important trick here is that sigma and integral signs can be changed around.

$$\int \sum^b_{n=a} f\left(n,x\right)\, dx = \sum^b_{n=a} \int f\left(n,x\right) \,dx$$

And this is because

$$\int \sum^b_{n=a} f(n,x)\, dx$$

$$\int f\left(a,x\right) + f((a+1),x) + f((a+2),x) + \dots +f\left((b-1),x\right) + f(b,x) $$

$$ = \int f(a,x)\,dx + \int f((a+1),x) \,dx + \dots + \int f((b-1),x)\, dx + \int f(b,x)\, dx$$


$$\begin{align*} \sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} =& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) \\ =& \sum_{n=0}^m \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\ =& \int_0^1 \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\ =& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x = \int_0^1 \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x \end{align*} $$

Also because $$ \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) = \frac{(1-x^{3(m+1)})(1-x)}{1-x^3} $$

Now let us see how the final integral

$$\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} $$

is evaluated.

$$ x^2+x+1 = \left(x+\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2$$

therefore if you skip two steps of substitution and do it once

$$ x+\frac{1}{2} = \frac{\sqrt{3}}{2} tan \theta$$

$$ dx = \frac{\sqrt{3}}{2} sec^{2} \theta$$

$$ \begin{eqnarray} \int \frac{dx}{1+x+x^2} = \int \frac{ \frac{\sqrt{3}}{2} sec^{2} \theta}{\frac{3}{4} sec^{2} \theta} {\mathrm{d} \theta} &=& \frac{2}{\sqrt{3}} \theta \\ &=& \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) \end{eqnarray} $$

$$ \Rightarrow \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \frac{2\sqrt{3}}{3} \left( tan^{-1} ( \frac{3}{\sqrt{3}} ) - tan^{-1} ( \frac{1}{\sqrt{3}} ) \right)$$

$$ = \frac{2\sqrt{3}}{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$$


Since it seems to have gone unnoticed I would like to mention another common trick that can be used here, namely to study the function $$f(z) = \frac{1}{(3z+1)(3z+2)} \pi\cot(\pi z)$$ which has the property that with $S$ being our sum, $$2S = \sum_n \mathrm{Res}(f(z); z=n) = \sum_n \frac{1}{(3n+1)(3n+2)}.$$ We examine what happens when we integrate $f(z)$ along the rectangle $$\Gamma =\pm (N+1/2) \pm i N$$ with $N$ a large positive integer. There are no poles on this contour and seeing that $\frac{1}{(3z+1)(3z+2)}\in\Theta(1/N^2)$ on the contour, the integral $$\int_\Gamma f(z) dz$$ goes to zero as $N$ goes to infinity.

This implies that the sum of the residues at the poles of $f(z)$ is zero, giving (observe the two simple poles at $z=-1/3$ and $z=-2/3$) $$2S + \frac{1}{3}\pi\cot(-\pi/3) - \frac{1}{3} \pi\cot(-2\pi/3) = 0$$ so that $$ S = \frac{\pi}{6} (\cot(-2\pi/3)-\cot(-\pi/3)) = \frac{\pi}{6} \frac{2}{\sqrt{3}} = \frac{\pi}{3\sqrt{3}}.$$

To convince yourself that the integral really does vanish consider the two lines $\Gamma_1$ which is $N+1/2\pm iN$ (right vertical) and $\Gamma_2$ which is $\pm N+1/2+iN$ (top horizontal).

We parameterize $\Gamma_1$ with $z=N+1/2+it$ so that $$\left|\int_{\Gamma_1} f(z) dz \right| = \left|\int_{-N}^N f(N+1/2+it) i dt\right|.$$ The fractional term attains its maximum at $t=0$ when we cross the real axis, giving an upper bound on the norm which is $$\frac{1}{(3N+5/2)(3N+7/2)}.$$ For the norm of the trigonometric term we get $$|\pi\cot(\pi (N+1/2) + \pi it)| =\pi\left|\frac{e^{i\pi (N+1/2) - \pi t}+e^{-i\pi (N+1/2) + \pi t}} {e^{i\pi (N+1/2) - \pi t}-e^{-i\pi (N+1/2) + \pi t}}\right| \\ = \pi\left|\frac{i(-1)^N e^{- \pi t} - i(-1)^N e^{\pi t}} {i(-1)^N e^{- \pi t} + i(-1)^N e^{\pi t}}\right| = \pi|\tanh(\pi t)|.$$ Observe that with $t$ real $\tanh(\pi t)$ has no poles and its norm is bounded by one. Therefore the norm of the integral along $\Gamma_1$ is bounded by $$2N \times \frac{1}{(3N+5/2)(3N+7/2)} \in \Theta(1/N)$$ and the integral vanishes as $N$ goes to infinity as claimed.

For $\Gamma_2$ we parameterize with $z = t + i N$ so that $$\left|\int_{\Gamma_2} f(z) dz \right| = \left|\int_{-(N+1/2)}^{N+1/2} f(t+iN) dt\right|.$$ The two factors of the fractional term are minimized when they cross the imaginary axis at $t=-1/3$ and $t=-2/3$, giving an upper bound on the norm which is $$\frac{1}{3N\times 3N} = \frac{1}{9N^2}.$$ For the norm of the trigonometric term we get $$|\pi\cot(\pi t + \pi i N)| = \pi\left|\frac{e^{i\pi t - \pi N} + e^{-i\pi t + \pi N}} {e^{i\pi t - \pi N} - e^{-i\pi t + \pi N}}\right| \le \pi\left|\frac{e^{\pi N}+e^{-\pi N}}{e^{\pi N}-e^{-\pi N}}\right| =\pi|\coth(\pi N)|.$$ There aren't any poles here either and this term is bounded above by $\pi\coth(\pi)$ because $N>1.$ This gives the following bound on the norm of the integral along $\Gamma_2:$ $$(2N+1) \times \frac{1}{9N^2} \times \pi\coth(\pi) \in \Theta(1/N)$$ and this integral also vanishes as $N$ goes to infinity as claimed.

The other two line segments can be bounded by the same technique.

  • $\begingroup$ I use the function $\pi\cot(\pi z)$ in this answer. However, avoiding contour integration made things simpler. $\endgroup$ – robjohn Jul 15 '14 at 23:57

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}} =\frac{1}{9}\sum_{n=0}^{\infty}\frac{1}{\pars{n + 1/3}\pars{n + 2/3}} =\frac{1}{3}\bracks{\Psi\pars{\frac{2}{3}} - \Psi\pars{\frac{1}{3}}} \end{align} where $\ds{\Psi}$ is the Digamma Function.

With Euler Reflection Formula: \begin{align}&\color{#66f}{\large% \sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}} =\frac{1}{3}\bracks{\pi\cot\pars{\pi\,\frac{1}{3}}} =\color{#66f}{\large\frac{\pi}{3\root{3}}} \end{align} because $\ds{\cot\pars{\frac{\pi}{3}} = {1 \over \root{3}}}$.


There is another way to solve the problem. Note $$ (3n+1)(3n+2)=9\left[(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2\right] $$ and and hence \begin{eqnarray} \sum_{n=0}^\infty\frac{1}{(3n+1)(3n+2)}&=&\frac19\sum_{n=0}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\ &=&\frac1{18}\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\ &=&\frac{1}{18}\frac{\pi\sinh2\pi b}{b(\cosh2\pi b-\cos2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{i}{6}}\\ &=&\frac{\pi}{3\sqrt3} \end{eqnarray} by using the result from this.


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