Exercise on Molien theorem Consider $G_m$ the mutiplicative group of $m$-roots of unity, acting on $\mathbb{C}[x,y]$ via $\epsilon\cdot(x,y)=(\epsilon^a x,\epsilon^{-a} y)$, where $0<a<m$ and $gcd(a,m)=1$.
I have to find the Poincarè-Hilbert series of the invariant ring,which I have already proven is $\mathbb{C}[x,y]^{G_m}=\mathbb{C}[x^m,xy,y^m]$; thus by Molien's theorem I know that the Poincarè-Hilbert series of the invariant ring is equal to $$HP_{G_m}(t)=\frac{1}{|M|}\sum_{\pi\in G_m} \frac{1}{\det(Id-t\pi)}.$$
Substituting stuff I obtain something of this form $$\frac{1}{m}\left[\frac{1}{(1-\epsilon^at)(1-\epsilon^{-a}t)}+\frac{1}{(1-\epsilon^{2a}t)(1-\epsilon^{-2a}t)}+\frac{1}{(1-\epsilon^{(m-1)a}t)(1-\epsilon^{-(m-1)a}t)}+\frac{1}{(1-t)^2} \right],$$
But at this point I'm quite stuck, I mean I've no idea how to simplify this term (and since my teacher did not suggest to use any software, I'm confident there's a clever way to simplify this and obtaining a prettier formula). Since this is my first exercise I'm afraid I have done some mistakes, that's why I'm asking it on the forum.
I even tried the simplest case ($m=3$, $a=1$), where at first I get
$$\frac{1}{3}\left[\frac{1}{(1-\epsilon t)(1-\epsilon^{-1}t)}+\frac{1}{(1-\epsilon^{2}t)(1-\epsilon^{-2}t)}+\frac{1}{(1-t)^2} \right],$$
but then since $\epsilon^2=\epsilon^{-1}$ I managed to obtain something like
$$\frac{1}{3}\left[\frac{2(1+t^2+2t)+(1-\epsilon t)^2(1-\epsilon^2 t)^2}{(1-\epsilon t)(1-\epsilon^{2}t)(1-t)^2} \right],$$
and I kept going without much hope.
Any hint, help or answer would be much appreciate, thanks in advance.
 A: Your question boils down to simplifying the expression
$$\frac{1}{m}\sum_{a=0}^{m-1}\frac{1}{(1-\epsilon^at)(1-\epsilon^{-a}t)}.$$
First and foremost it is worth noting that in general
$$\frac{1}{1-x}=\sum_{n\geq0}x^n,$$
enabling you to turn the sum of reciprocals into a proper power series in $t$. We get
\begin{eqnarray*}
\frac{1}{(1-\epsilon^at)(1-\epsilon^{-a}t)}
&=&\left(\sum_{n\geq0}(\epsilon^at)^n\right)\cdot\left(\sum_{n\geq0}(\epsilon^{-a}t)^n\right)\\
&=&\sum_{n\geq0}\left(\sum_{k=0}^n(\epsilon^a)^k(\epsilon^{-a})^{n-k}\right)t^n\\
&=&\sum_{n\geq0}\left(\sum_{k=0}^n(\epsilon^a)^{2k-n}\right)t^n.
\end{eqnarray*}
Now summing over all values of $a$ we find that
\begin{eqnarray*}
\sum_{a=0}^{m-1}\frac{1}{(1-\epsilon^at)(1-\epsilon^{-a}t)}
&=&\sum_{n\geq0}\left(\sum_{k=0}^n\sum_{a=0}^{m-1}(e^a)^{2k-n}\right)t^n.\\
&=&\sum_{n\geq0}\left(\sum_{k=0}^n\sum_{a=0}^{m-1}(e^{2k-n})^a\right)t^n.
\end{eqnarray*}
These inner sums have a nice closed form; it is not a difficult exercise to show that
$$\sum_{a=0}^{m-1}(\epsilon^{2k-n})^a=\begin{cases}
0&\text{ if }m\nmid(2k-n)\\
m&\text{ if }m\mid(2k-n)
\end{cases}.$$
And so the original expression greatly simplifies to
$$\frac1m\sum_{a=0}^{m-1}\frac{1}{(1-\epsilon^at)(1-\epsilon^{-a}t)}=\frac1m\sum_{n\geq0}mt^{mn}=\frac{1}{1-t^m}.$$
