Find the value of $\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}$ I was doing some problems in algebraic number theory and this series came up
$$\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}.$$
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.
Thanks
 A: 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:
$$\frac{p(z)}{1-z^d}=-\frac{1}{d}\sum_{k=0}^{d-1}\frac{p(\zeta_d^i)\zeta_d^i}{z-\zeta_d^i}$$
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.
A: 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
$$\sum_{n=0}^{\infty} \frac1{9 n^2+9 n+2} = \frac{\pi}{3 \sqrt{3}} $$
A: 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 
$$\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}=\frac{4}{9}\sum_{n=0}^\infty\frac{1}{(2n+1)^2-\frac{1}{9}}=\frac{4}{9}\frac{\pi}{4\left(\frac{1}{3}\right)}\tan\left(\frac{\pi }{6}\right)=\frac{\pi}{3\sqrt{3}}$$
A: 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}
