Interesting limit with Poisson and Chi-squared Distribution I am stuck with computing the following limit: 
\begin{align}
 \lim_{ n \to \infty} E \left[ \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U  \right] \right)^2 \right].
\end{align} 
In the above expression,  $X$ given $U$ follows Poisson with parameter $U$  and where $U$ is a Chi-square of degree $n$. 
Here is what I tried: 
Suppose we let
\begin{align}
V_n =\left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2
\end{align} 
Then, using Jensen's inequality
\begin{align}
V_n  \le E \left[  \frac{X}{n} + \frac{1}{2} \,  \Big | U \,  \right] =  \frac{U}{n}+\frac{1}{2}
\end{align} 
Moreover, we have that $E \left[  \frac{U}{n}+\frac{1}{2}   \right]=1+\frac{1}{2}$. Therefore, by the dominated convergence theorem we have that 
\begin{align}
\lim_{n \to \infty}  E[  V_n ]=   E[   \lim_{n \to \infty}  V_n ]
\end{align}
Therefore, assuming everything up to here is correct, to compute the limit we have to  find 
\begin{align}
 \lim_{n \to \infty}  V_n&=  \lim_{n \to \infty}   \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2\\
&=  \left(  \lim_{n \to \infty}  E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}}  \, \Big | \, U  \right] \right)^2
\end{align} 
This is the place where I am stuck.  Is it simply another applications dominated convergence theorem?  If so, then I think the answer is
\begin{align}
 \lim_{ n \to \infty} E \left[ \left(E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U  \right] \right)^2 \right]=\frac{1}{2}.
\end{align} 
What I mean by another application of dominating convergence theorem is that for every $u>0$
\begin{align}
E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right] &\le  \sqrt{ E \left[   \frac{X}{n} + \frac{1}{2} \, \Big  | \, U=u  \right]}\\
&=   \sqrt{  \frac{u}{n} + \frac{1}{2} }\\
&= \sqrt{  u + \frac{1}{2}}
\end{align}
Therefore, 
\begin{align}
E \left[  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right]=  E \left[  \lim_{n \to \infty}  \sqrt{ \frac{X}{n} + \frac{1}{2}} \, \Big  | \, U=u  \right]=  \frac{1}{2}.
\end{align}
Is this a correct sequence of steps?  I feel a bit uneasy about the second application of the dominated convergence theorem. 
 A: proof: Instead of the heuristic argument I made below you can simply prove it by finding a lower bound, since you already found an upper bound of $3/2$ by Jensen's inequality. Using
$$\sqrt{1+2x}\geq \sqrt{3} \left(1+\frac{x-1}{3}\right) - (x-1)^2$$
$\forall x \geq 0$, which then becomes with $x=\frac{X}{n}$
$$\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right] \geq \mathbb{E}\left[ \sqrt{3} \left(1+\frac{X/n-1}{3}\right) - (X/n-1)^2 \right] \\
=-{\frac {{U}^{2}}{{n}^{2}}}+ \left( -\frac{1}{n^2} + \frac{1}{\sqrt{3}\,n} + \frac{2}{n} \right) U + \frac{2}{\sqrt{3}}-1$$
and so since
$$\mathbb{E}\left(U\right) = n \\ \mathbb{E}\left(U^2\right) = n(n+2)$$
then
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right] \geq \sqrt{3}$$
which means
$$\lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]^2\right] \geq \lim_{n\rightarrow \infty} \mathbb{E} \left[\mathbb{E}\left[ \sqrt{1+\frac{2X}{n}} \right]\right]^2 \geq \sqrt{3}^2 = 3 \tag{FKG}$$
because for any $u$ from the domain of the chi-squared distribution and since $\sqrt{1+2x}$ is increasing we have
$$\frac{{\rm d}}{{\rm d}u} \sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\sqrt{1+2k/n}=\sum_{k=0}^\infty \frac{u^k {\rm e}^{-u}}{k!}\left(\sqrt{1+2(k+1)/n}-\sqrt{1+2k/n}\right)\geq 0$$ and so $\mathbb{E}\left[\sqrt{1+\frac{2X}{n}}\right]$ is an increasing function in $U=u$ and the reqiurements for the FKG inequality are fulfilled. $\square$

Originally I made a too long of a comment for a heuristic argument. Though I know I'm completely ignoring convergence here, when you simply manipulate the corresponding series expressions you obtain the Jensen upper bound result of $3/2$. Starting with
$$
\sqrt{1+\frac{2X}{n}}=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2X}{n}\right)^m
$$
and then calculating ${\mathbb{E}(\cdot)}$ with respect to Poisson one obtains
$$
\mathbb{E}\left(\sqrt{1+\frac{2X}{n}}\right)=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \mathbb{E}\left(X^m\right) \\
=\sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m \sum_{i=0}^m {m \brace i} \, U^i \\
=\sum_{i=0}^\infty U^i \sum_{m=i}^{\infty} \binom{1/2}{m} \left(\frac{2}{n}\right)^m {m \brace i} \\
\stackrel{m=i+k}{=}\sum_{i=0}^\infty \left(\frac{2U}{n}\right)^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i} \, .
$$
The $i$-th moment with respect to the chi-squared distribution of degree $n$ is $$\mathbb{E}\left(U^i\right) = \frac{2^i \Gamma(i+n/2)}{\Gamma(n/2)} \sim n^i \left(1+\frac{i(i-1)}{n} + {\cal O}(1/n^2) \right)$$
which is merely an asymptotic series, but taking only the first term, squaring the previous equation and taking $\mathbb{E}(\cdot)$ we obtain
$$
\left(\sum_{i=0}^\infty 2^i \sum_{k=0}^{\infty} \binom{1/2}{i+k} \left(\frac{2}{n}\right)^k {i+k \brace i}\right)^2 \, .
$$
The terms $k>0$ vanish for $n\rightarrow \infty$, so we are left with
$$
\left( \sum_{i=0}^\infty \binom{1/2}{i} \, 2^i \right)^2 = \sqrt{1+2}^2 = 3
$$
where the series again doesn't converge, but only in the sense of analyticity. Gathering the left out overall factor $1/2$ the final result yields $3/2$.
A: Let $\xi_1,\xi_2,\ldots$ be i.i.d. $N(0,1)$. Let $N_1,N_2,\ldots$ be i.i.d., unit rate Poisson processes, constructed so that $\{N_j\}$ and $\{\xi_j\}$ are independent.
Define $Y_j=N_j(\xi_j^2)$. Then $Y_1,Y_2,\ldots$ are i.i.d. and the conditional distribution of $Y_j$ given $\xi=\{\xi_1,\xi_2,...\}$ is Poisson with parameter $\xi_j^2$.
Let $X_n=Y_1+\cdots+Y_n$ and $U_n=\xi_1^2+\cdots+\xi_n^2$. Then $U_n$ has a chi-squared distribution with $n$ degrees of freedom and the conditional distribution of $X_n$ given $\xi$ is Poisson with parameter $U_n$. We are interested in $\lim_{n\to\infty}EV_n$, where $V_n=(E[Z_n\mid U_n])^2$ and
$$
Z_n = \sqrt{\frac{X_n}n + \frac12}.
$$
Note that $V_n=(E[Z_n\mid\xi])^2$.
By the law of large numbers, $X_n/n\to EY_1=E\xi_1^2=1$ a.s. Thus, $Z_n\to\sqrt{3/2}$ a.s. Since $EZ_n^2=3/2$ for all $n$, it follows that $\{Z_n\}$ is uniformly integrable. Therefore, $Z_n\to\sqrt{3/2}$ in $L^1$, which implies that $E[Z_n\mid\xi]\to\sqrt{3/2}$ in $L^1$.
By passing to a subsequence, we may assume this convergence is almost sure. Then, along this subsequence, $V_n\to3/2$ a.s. Now,
\begin{align}
EV_n^2 &\le EZ_n^4\\
&= E\left|{\frac{X_n}n + \frac12}\right|^2\\
&= n^{-2}EX_n^2 + n^{-1}EX_n + \frac14\\
&= n^{-2}EX_n^2 + 1 + \frac14.
\end{align}
Since
\begin{align}
EX_n^2 &= E[E[X_n^2\mid U_n]]\\
&= E[U_n + U_n^2]\\
&= n + n(n+1)\\
&= n^2 + 2n,
\end{align}
we have
$$
\sup_nEV_n^2
\le \sup_n\left({1 + \frac2n + 1 + \frac14}\right)
< \infty.
$$
Hence, $\{V_n\}$ is uniformly integrable, and so $EV_n\to3/2$ along the given subsequence.
This shows that every subsequence of $\{EV_n\}$ has a further subsequence converging to $3/2$. Therefore $EV_n\to3/2$.
EDIT:
Here is an alternative to the second half of the proof which uses a modification of your dominated convergence idea.
Follow the proof up until the point where we show that $V_n\to3/2$ a.s. Now,
$$
V_n \le W_n := E\left[{
  \frac{X_n}n + \frac12 \;\bigg|\; U_n
}\right]
= \frac{U_n}n + \frac12.
$$
By the law of large numbers, $W_n\to3/2$ a.s. Also, $EW_n=3/2$ for all $n$. Thus, by the generalized dominated convergence theorem (see General Lebesgue Dominated Convergence Theorem, for example), we have $EV_n\to 3/2$.
EDIT 2:
In fact, by mimicking the proof of the generalized DCT (as given in the above link) and using Fatou's lemma for conditional expectation (Theorem 6.56 in these notes), you can prove the following:

Theorem. Suppose $X_n\to X$ a.s., $|X_n|\le Y_n$, $Y_n\to Y$ a.s., and $E[Y_n\mid\mathcal{G}]\to E[Y\mid\mathcal{G}]$. Then $E[X_n\mid\mathcal{G}]\to E[X\mid\mathcal{G}]$ a.s.

We can now use this theorem to give an alternative proof that $V_n\to3/2$ a.s. First, $Z_n\to\sqrt{3/2}$ a.s. by the law of large numbers. Next, since $x\ge1/2$ implies $\sqrt x\,\le\sqrt 2\,x$, we have
$$
Z_n \le \zeta_n := \sqrt2\,\left({\frac{X_n}n + \frac12}\right).
$$
Since $\zeta_n\to3/\sqrt2$ a.s. and $E\zeta_n=3/\sqrt2$ for all $n$, the above theorem implies
$$
E[Z_n\mid U_n] = E[Z_n\mid\xi]\to\sqrt{3/2}\quad\text{a.s.}
$$
If you use both these alternates, then you have a rigorous incarnation of what you originally envisioned, which was two applications of the dominated convergence theorem.
