Find the value of $\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}$

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 :)

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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.

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

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