In came across the result that the expectation of the Fisher score function is zero so I wanted to check this for a discrete distribution (in my case Poisson) since my textbook only focusses on continuous distributions (although the discrete case should be similar).
To this end I first computed the log-likelihood function $l(\lambda,y)$ which for the Poisson distribution is given by: $l(\lambda,y) = \sum_{i=1}^{n} [y_i] \cdot \ln(\lambda) - n\lambda - \ln[\prod_{i=1}^n [y_i!]]$ and consequently the score function which is given by: $u(\lambda,y) = \frac{\sum_{i=1}^{n} [y_i] - n\lambda}{\lambda}$.
Now $E[u(\lambda,Y);\lambda] = 0$ should hold. During the process of showing this result however I get stuck in the summations. I have performed the following steps:
$E[u(\lambda,Y);\lambda] = \sum_{i=0}^{\infty} [\frac{\sum_{i=1}^{n} [y_i] - n\lambda}{\lambda} \cdot \frac {\lambda^{y_i} \cdot e^{-\lambda}}{y_i!}] = \frac{e^{-\lambda}}{\lambda} \cdot (\sum_{i=0}^{\infty} [\frac{\sum_{i=1}^{n} [y_i] \cdot \lambda^{y_i} }{y_i!}] - n\lambda \cdot \sum_{i=0}^{\infty} [\frac{\lambda^{y_i}}{y_i!}]) = \ \ \ \ \frac{e^{-\lambda}}{\lambda} \cdot (\sum_{i=0}^{\infty} [\frac{\sum_{i=1}^{n} [y_i] \cdot \lambda^{y_i} }{y_i!}] - n\lambda \cdot e^{\lambda})$
(in the last step I use the power series expansion of $e^{\lambda}$).
Is the above correct? And if so, how should I proceed?