Let $N$ be Poisson with parameter $m$. Find $\frac{E[ \log (N+\frac{1}{m} )]}{\log(m)}$ Let $N$ a Poisson random variable with parameter $m$.  How to find the following limit:
\begin{align}
\lim_{m \to \infty} \frac{E[\log(N+\frac{1}{m})]}{\log(m)} ? 
\end{align}
Using Jensen's inequality we have that
\begin{align}
\lim_{m \to \infty} \frac{E[\log(N+\frac{1}{m})]}{\log(m)} \le\lim_{m \to \infty} \frac{\log(E[ N+\frac{1}{m}])}{\log(m)}= \lim_{m \to \infty} \frac{\log(m+\frac{1}{m})}{\log(m)}=1.
\end{align}
I am wondering if this is the correct behavior.
 A: Indeed it is, note that $$ \ln(N+\frac{1}{m})  = \ln(\frac{N+\frac{1}{m}}{m}) + \ln(m)$$ so $$ \mathbb E[\frac{\ln(N+\frac{1}{m})}{\ln(m)}] =  = \frac{1}{\ln(m)}\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})] + 1 $$
We need to prove that $$ \frac{1}{\ln(m)}\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})]$$ converges to $0$. As $N \sim \mathcal Po(m)$ we get (second one we bounded by $ln(x) \le x$ for $x>0$) $$ |\mathbb E[\ln(\frac{N+\frac{1}{m}}{m})]| \le |e^{-m}\sum_{k=0}^m \frac{m^k}{k!}\ln(\frac{k+\frac{1}{m}}{m})| +  e^{-m} \sum_{k=m}^\infty \frac{m^{k-1}}{k!}(k+\frac{1}{m}) $$
We can split the second series into $2$, namely $$ e^{-m}\sum_{k=m}^\infty \frac{m^{k-1}}{(k-1)!} + e^{-m} \frac{1}{m} \sum_{k=m}^\infty \frac{m^{k-1}}{k!} $$
And it is not hard to see that both series are bounded by $e^m$, hence the whole series is bounded by something convering to $1$ and by multiplying it by $\frac{1}{\ln(m)}$ we have convergence to $0$.
The first $m$ terms are harder. Split the sum into $k \in \{0,...,\frac{m}{2}\}$ and $k \in \{\frac{m}{2} , ... , m\}$. The second one can be bounded by $$ | \sum_{k=\frac{m}{2}}^m e^{-m}\frac{m^k}{k!} \ln(\frac{1}{2} + \frac{1}{m^2})| \le |\ln(\frac{1}{2} + \frac{1}{m^2}|$$ which converges to $0$ if we multiply it by $\frac{1}{\ln(m)}$.
While the second one requires either analitic computation or central limit theorem. Indeed, note $$ \frac{1}{\ln(m)}|\sum_{k=0}^{\frac{m}{2}} e^{-m}\frac{m^k}{k!}\ln(\frac{k}{m} + \frac{1}{m^2})| \le \frac{1}{\ln(m)}|\ln(\frac{1}{m^2}) | \sum_{k=0}^{\frac{m}{2}} e^{-m}\frac{m^k}{k!}  = 2 \mathbb P(N \le \frac{m}{2})$$
It remains to show that $\mathbb P(N \le \frac{m}{2})$ converges to $0$, but as $N$ is sum of $n$ independent $\mathcal Po(1)$ random variables (call them $X_k$) we have due to central limit theorem $$ \mathbb P(N \le \frac{m}{2}) = \mathbb P( \sum_{k=1}^m X_k - m \le -\frac{m}{2}) = \mathbb P( \frac{\sum_{k=1}^m (X_k-\mathbb E[X_k])}{\sqrt{m}} \le -\frac{\sqrt{m}}{2}) \to \mathbb P(\mathcal N(0,1) \le -\infty) = 0$$
W which proves our claim.
