Coefficients of a Zero Forcing linear equalizer for infinite length input
The following are hints as this is reads as a homework question. Please note that these are not ground breaking hints and can be found in a good text or summary on zero forcing somewhere I'm sure.
Hint 1
The first step would be to look at the model and be able to pick out what the channel is (you have channel convolved with signal plus noise). It seems you have identified the channel as, $h_k = \alpha^{-k}u_k$, where $u_k$ is the unit step function, and the z-transform can be found by directly applying the formula.
\begin{align}
H(z) &amp;= \sum_{k=-\infty}^{\infty}h_k z^{-k} \\
&amp;= \sum_{k=-\infty}^{\infty} \big(\alpha^{-k}u_k \big)z^{-k} \\
&amp;= \sum_{k=0}^{\infty} \alpha^{-k} z^{-k} \\
&amp;= \alpha^0z^0+\alpha^{-1}z^{-1}+\alpha^{-2}z^{-2}+\cdots
\end{align}
Just be careful when simplifying...
Hint 2
Zero forcing (ZF) means that the equalizer is trying to completely eliminate intersymbol interference (ISI), and ZF does achieve this in the noise free case. For the noise free case, in the z-domain your model becomes:
$$ Y(z) = H(z)A(z) $$
The ZF equalizer should be $W_{\text{ZF}}(z)=\frac{1}{H(z)}$ so that the equalized signal will have eliminated the ISI caused by the channel.
Hint 3
This should become clear when you consider what happens to each symbol as it goes through the channel. Take a single symbol being sent over the channel at $k=0$. The received noise free version is $a_0+\alpha^{-1}a_{-1}+\cdots$, or in words this is, $\text{Original symbol } + \text{ Previous symbol } + \cdots$. Can you pick an $\alpha$ to ensure the only non-zero value is the original symbol?