I am frustrated with why this is happening. The distribution is truncated Poisson with $\theta$ and $y_1,...,y_n$ observations.
If $l(\theta)$ denotes the log likelihood of the distribution, I need to find(skipped tedious calculations to get to this point but the $l''$ is correct)
$$-\mathbb{E}[l''(\theta)]=-\mathbb{E}[\frac{-n\bar{y}}{\theta^2}+n\frac{e^{-\theta}}{(1-e^{-\theta})^2}]$$
Well, here are my steps
$-\mathbb{E}[\frac{-n\bar{y}}{\theta^2}+n\frac{e^{-\theta}}{(1-e^{-\theta})^2}]=-\mathbb{E}[\frac{-n\bar{y}}{\theta^2}]-\mathbb{E}[n\frac{e^{-\theta}}{(1-e^{-\theta})^2}]=\frac{n}{\theta^2}\mathbb{E}[\bar{y}]-n\frac{e^{-\theta}}{(1-e^{-\theta})^2}$
$\bar{y}=\frac{\Sigma y_i}{n}$ so therefore,
$\frac{n}{\theta^2}\mathbb{E}[\bar{y}]-n\frac{e^{-\theta}}{(1-e^{-\theta})^2}=\frac{1}{\theta^2}\mathbb{E}[\Sigma y_i]-n\frac{e^{-\theta}}{(1-e^{-\theta})^2}=\frac{1}{\theta^2}n\theta-n\frac{e^{-\theta}}{(1-e^{-\theta})^2}$ since $y_i$ is distributed with truncated Poisson that has mean $\theta$.
Thus $\frac{n}{\theta}-n\frac{e^{-\theta}}{(1-e^{-\theta})^2}$
However, the solution says the answer is
$$n(\frac{1}{\theta(1-e^{-\theta})}-\frac{e^{-\theta}}{(1-e^{-\theta})^2})=\frac{n(1-e^{-\theta}-\theta e^{-\theta})}{\theta(1-e^{-\theta})^2}$$
Not convinced. Especially because it skips the calculation steps. I have been staring it for an hour, retried the calculation and no, I don't see a flaw in my steps. I need someone more experienced to point it out. I seriously cannot see why the answer got this form, in mine, I clearly cannot avoid having $(1-e^{-\theta})^2$ to calculate the fractions but the answer somehow manages to avoid that.