Convergence of poisson distribution Let $X\sim \operatorname{Pois}(\lambda)$ and $x_1,\ldots,x_n$ observations following this distribution.
I want to derive the analytical solution of the following series:
$$\ell(\lambda):=\lim_{x\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n \log P(X=x_i).$$
EDIT: After a few trials, I found a good numerical approximation of the solution:
$$\ell(\lambda)=-\frac{1}{2}\log(17.08\cdot \lambda).$$
See the graph below, where the dots represent an approximation of the solution by simulating poisson distributions, while the blue line represents the approximated numerical solution.
x=1:1000
y=sapply(x,function(x) mean(log(dpois(rpois(100000,x),x))))
plot(x,y)
lines(x,-log(sqrt(x*17.08)),col="blue")


 A: If you have $n$ samples of a random variables $Y$  (or equivalently, $n$ iid random variables) and you compute the sample mean as
$$ m_n = \frac{1}{n} \sum_{i=1}^n Y_i$$
then, as $n$ grows, (under certain conditions) $m_n$ will converge (in some sense) to the expected value of $Y$, that is $m_n \to E[Y]$. See the law of large numbers, which clarifies those "certain conditions" and in what sense the "convergence" occurs.
In our case, $Y=\log(p(X))$ hence
$$\begin{align}
 m_n \to E[\log(p(X)]&=\sum_x p_X(x) \log(p_X(x)) \\
&= \sum_{x=0}^\infty  e^{-\lambda} \frac{\lambda^x}{x!} (-\lambda + x \log(\lambda) - \log(x!)) \\
&= -\lambda (1 - \log(\lambda)) - e^{-\lambda} \sum_{x=0}^\infty\frac{\lambda^x}{x!}\log(x!)
\end{align}$$
This cannot be put in simpler form, but for large $\lambda$ it can be shown that it tends to $ -\frac{1}{2} \log(2 \pi e \lambda)$
Except for the sign, as noted in a comment, this is the entropy of a Poisson distribution, see eg here or the Wikipedia.
BTW, if you look for higher order correction terms, bear in mind that most results in the bibliography imply a base two logarithm (as usual in information theory).
