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

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 $-1\leq x \leq1$...

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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. – 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: $$\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2 \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}}$$

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 Clever. The actual computations are probably not so terribly different from my idea. – Harald Hanche-Olsen Feb 22 '12 at 21:26 Proceeding per your suggestion leads to more complicated integrals: $\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \mathrm{d} x \int_0^x \mathrm{d} y \sum_{n=0}^\infty y^{3n} = \int_0^1 \mathrm{d} x \int_0^x \frac{\mathrm{d} y }{1-y^3}$. – Sasha Feb 22 '12 at 21:42 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. – Harald Hanche-Olsen Feb 22 '12 at 21:46 Thank you for this nice answer! However woundn't it rather be: $$\int_0^1 \frac{1-x^{3(m+1)}}{1+x + x^2} \mathrm{d} x$$ ? – Chon Feb 23 '12 at 7:53 Yes, you are correct. I will edit the post. – 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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Why do I like answering a question with a question? – Harald Hanche-Olsen Feb 22 '12 at 21:23
Why not? [And now the system makes me spoil the joke by forcing me to type more.] – André Nicolas Feb 22 '12 at 21:36
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.] – Harald Hanche-Olsen Feb 22 '12 at 21:43
Type \ \ \ \ \ \ between dollar signs. – David Mitra Feb 22 '12 at 21:56

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

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

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

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

Therefore

\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

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

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

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

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

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

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