<s>[INST] Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$ Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]. 
\end{align}
Things that I tried:
First, I found an upper bound by using Jensen's inequality
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le  \lim_{m \to \infty} m  \log\left( \frac{ E \left[ N  \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m  \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}. 
\end{align}
For the lower bound I tried to use the CLT argument from a related question in here:
\begin{align}
m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ]
\end{align}
for some $k>0$.  Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that
\begin{align}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] .
\end{align}
Now if we use the lower bound at this point, we get
\begin{align*}
 \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right]  \ge   \lim_{m\to \infty}m  \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty,
\end{align*}
for all $k \in (0,1)$.
 [/INST] We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.
We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$
and thus
$$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+\frac12}  -\frac12.$$
It's sufficient to show that $\lim_m E\ln \frac{N+ \frac12}{m+\frac12} \to -\frac{1}2$.
Condsider $m \ge 100$, $1+Y_i \sim Pois(1)$, $\xi_m \sim Pois(m - [m])$. Thus $\sum_{i=1}^{[m]} (Y_i + 1) + \xi_m = N$ in distribution.
Fix $x$. $$P(\frac{\xi_m}{\sqrt{m}} \ge x) \le \frac{E \frac{\xi_m}{\sqrt{m}} }{x} \to 0, m \to \infty$$
thus $ \frac{\xi_m}{\sqrt{m}}  \to 0$ in probability.
It follows from CLT that $\frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}} \to N(0,1)$. Hence
$$\frac{N-m}{\sqrt{m}} = \frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}}\frac{\sqrt{[m]}}{\sqrt{m}}   +  \frac{\xi_m}{\sqrt{m}} \to N(0,1).$$
Put $$\eta_m = \frac{m}{m+\frac12}\cdot \frac{N-m}{\sqrt{m}} = \frac{\sqrt{m}(N-m)}{m+\frac12}$$
Then $\eta_m \to N(0,1)$  and $\eta_m \ge \frac{m}{m+\frac12}\frac{-m}{\sqrt{m}} \ge -\sqrt{m} + \frac{1}{3\sqrt{m}}$.
Further,
$$\lim_m E\ln \frac{N+ \frac12}{m+\frac12} = E m \ln(1 + \frac{\eta_m}{\sqrt{m}}).$$
Put $g(x) = \ln(1+x) - (x - \frac{x^2}2)$. Then
$$ E m \ln(1 + \frac{\eta_m}{\sqrt{m}}) = E m(  \frac{\eta_m}{\sqrt{m}} - \frac12 ( \frac{\eta_m}{\sqrt{m}})^2 + g( \frac{\eta_m}{\sqrt{m}}) ) = - \frac12 + Emg( \frac{\eta_m}{\sqrt{m}})  $$
and it's sufficient to show that $Emg( \frac{\eta_m}{\sqrt{m}})  \to 0$.
Put $A = [-\sqrt{m}+ \frac{1}{3\sqrt{m}}, - m^{\frac{3}{8}}], B= [-m^{\frac{3}{8}}, m^{\frac{3}{8}}], C = [m^{\frac{3}{8}}, \infty)$. Then
$$|Emg( \frac{\eta_m}{\sqrt{m}})| \le Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le $$
$$\le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in A)  + m \max_{x \in B} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in B) + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C}$$
As $m\ge 100$, we know that  $\max_{x \in A} |g(\frac{x}{\sqrt{m}})| = |\ln(\frac{1}{3m})| \le 2 \ln m $.
Further, by Taylor formula
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$, $0 \le \theta \le 1$. Thus
$g(x) = \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$
and $\max_{x \in B} |g(\frac{x}{\sqrt{m}})| =  O(\frac{1}{ m^{\frac98 }  })$.
We have $|g(x)| \le 2x^2$ for $x > 0$.
Thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le E m 2 (\frac{\eta_m}{\sqrt{m}})^2 I_C = 2 E \eta_m^2 I_C$.
Further there will be heuristic arguments.
$2 E \eta_m^2 I_C \approx 2 E N^2(0,1) I_{(N(0,1) > m^{\frac38})} \to 0$ hence $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \to 0$.
$P(\eta_m \in A) \approx P(N(0,1) \in A) \le P(N(0,1) \le -m^{\frac38})$ and thus
$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A}  \le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|  P(N(0,1) \le -m^{\frac38}) \to 0$.
It's sufficient to show that$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \to 0$.
But $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \le m \max_{x \in B} |g(\frac{x}{\sqrt{m}})| = m  O(\frac{1}{ m^{\frac98 }  }) = o(1)$, q.e.d.
</s>