I was doing some problems in algebraic number theory and this series came up


So, I would like to know the value of this series. However, I don't want a full answer, only hints. I have tried to use Fourier's series, but with no success, perhaps an answer can be gotten usign this.


  • 4
    $\begingroup$ Hint: $\dfrac{1}{9n^2+9n+2}=\dfrac{1}{(3n+1)(3n+2)}=\dfrac{1}{3n+1}-\dfrac{1}{3n+2}$ $\endgroup$ – Amir Naseri Jan 29 '16 at 18:22
  • $\begingroup$ $9n^2+9n+2=(3n+1)(3n+2)$ so you can make this an alternating sequence. Not sure how much that helps. $\endgroup$ – Thomas Andrews Jan 29 '16 at 18:23
  • 1
    $\begingroup$ A related question. $\endgroup$ – Lucian Jan 29 '16 at 22:14

Let $f(z)=\frac{1}{1+z+z^2}=\frac{1-z}{1-z^3}=1-z+z^3-z^4+z^6-z^7\dots$.

Then this is the derivative of $$\frac{z}{1}-\frac{z^2}{2}+\frac{z^4}{4}-\frac{z^5}{5}+\cdots$$.

Now, the anti-derivative of $\frac{1}{1+z+z^2}=\frac{4}{(2z+1)^2+3}$. Can be computed, and then take the limit as $z\to 1$.

Wolfram gives the anti-derivative as:

$$ \frac{2 \tan^{-1}((2 z+1)/\sqrt{3})}{\sqrt{3}}$$

[Not a hard integral to do by hand, just not feeling able to do an integral at the moment.]

As $z\to 1$ this gives $\frac{\pi}{3\sqrt{3}}$.

Why can we let $z\to 1$ to get the value? Mumble mumble something mumble analytic function, mumble, then a miracle occurred. I forget why we can do this, but we definitely have to know the series converges to do this :)

This next step is a work on progress.

More generally if $p(z)=a_0+a_1z+\cdots a_{d-1}z^{d-1}$ is a polynomial then:


where $\zeta_d$ is a primitive $d$th root of $1$. The anti-derivative will be in terms of complex logarithms, and won't converge as $z\to 1$ unless $p(1)=0$.

Then the anti-derivative at $z=1$ will be $$-\frac{1}{d}\sum_{i=1}^{d-1} p(\zeta_d^i)\zeta_d^i \log(1-\zeta_d^i)$$

Since we are looking at complex logarithms, that's going to involve $\arctan$ and the natural logarithm.

  • $\begingroup$ Well, we can easily compare it to $\frac{1}{n^2}$ to show convergence $\endgroup$ – zz20s Jan 29 '16 at 18:49
  • $\begingroup$ Oh, I know we know convergence, I was just fudging why that is enough. :) @zz20s $\endgroup$ – Thomas Andrews Jan 29 '16 at 18:54

You can use the residue theorem. Rewrite the sum as

$$\frac1{18} \sum_{n=-\infty}^{\infty} \frac1{\left (n+\frac13 \right )\left (n+\frac23 \right )} $$

Then use the fact that a consequence of the residue theorem is

$$\begin{align}\sum_{n=-\infty}^{\infty} \frac1{\left (n+\frac13 \right )\left (n+\frac23 \right )} &= -\pi \sum_k \operatorname*{Res}_{z=z_k} \frac{\cot{\pi z}}{\left (z+\frac13 \right )\left (z+\frac23 \right )} \\ &= \pi \left [\frac{\cot{(\pi/3)}}{1/3} - \frac{\cot{(2 \pi/3)}}{1/3}\right ]\\ &= 2 \sqrt{3} \pi\end{align}$$

Therefore the sum is

$$\sum_{n=0}^{\infty} \frac1{9 n^2+9 n+2} = \frac{\pi}{3 \sqrt{3}} $$

  • $\begingroup$ Awfully sophisticated proof for a simple fact haha $\endgroup$ – ParaH2 Jan 29 '16 at 18:44

we know that $$\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2-x^2}=\frac{\pi}{4x}\tan\left(\frac{\pi x}{2}\right)$$ so $$\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}=\frac{4}{9}\sum_{n=0}^\infty\frac{1}{(2n+1)^2-\frac{1}{9}}=\frac{4}{9}\frac{\pi}{4\left(\frac{1}{3}\right)}\tan\left(\frac{\pi }{6}\right)=\frac{\pi}{3\sqrt{3}}$$

  • $\begingroup$ Nice solution, but the OP asked for hints only. $\endgroup$ – rogerl Jan 29 '16 at 20:33

Note that $$ \sum_{n=0}^{\infty}\frac{1}{9n^2+9n+2}x^{3n}=\sum_{n=0}^{\infty}\left[\frac{1}{3n+1}x^{3n}-\frac{1}{3n+2}x^{3n}\right]. $$ Define $$ f(x)=\sum_{n=0}^{\infty}\frac{1}{3n+1}x^{3n}, g(x)=\sum_{n=0}^{\infty}\frac{1}{3n+2}x^{3n}. $$ It is easy to check that $$ (xf(x))'=\frac{1}{1-x^3}, (x^2g(x))'=\frac{x}{1-x^3}. $$ So

\begin{eqnarray} \lim_{x\to1}f(x)-g(x)&=&\lim_{x\to1}\frac1x\int_0^X\frac{1}{1-t^3}dt-\frac1{x^2}\int_0^x\frac{t}{1-t^3}dt\\ &=&\lim_{x\to1}\int_0^1\frac{1-u}{1-(ux)^3}du\\ &=&\int_0^1\frac{1-u}{1-u^3}du\\ &=&\int_0^1\frac{1}{1+u+u^2}du\\ &=&\frac{\pi}{3\sqrt3} \end{eqnarray}



$$\sum_{n=0}^{\infty}\frac{1}{9n^2+9n+2}=\sum_{n=0}^{\infty}\left[\frac{1}{3n+1}-\frac{1}{3n+2}\right]=$$ $$\lim_{m\to\infty}\left[\sum_{n=0}^{m}\frac{1}{3n+1}-\sum_{n=0}^{m}\frac{1}{3n+2}\right]=\lim_{m\to\infty}\frac{1}{9}\left[\pi\sqrt{3}+3\text{H}_{\frac{1}{3}+m}-3\text{H}_{\frac{2}{3}+m}\right]$$


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