Setup
r.v. $X_1, \ldots, X_n \sim i.i.d. Po(\lambda)$. Now I consider a testing $H_0: \lambda = \lambda_0, H_1: \lambda > \lambda_0$.
This is Example 3.3.2 of "Elements of Large-Sample Theory" (p.162).
Question
I need to show that \begin{align} \frac{\sqrt{n}(\bar{X}-\lambda_n)}{\sqrt{\lambda_n}} \xrightarrow{L} N(0,1)\ \ \ \mathrm{as}\ \lambda_n \to \lambda_0. \end{align}
What I know
$$ \frac{\mathbb{E}[|X_1 - \lambda_n|^3]}{\lambda_n^{3/2}} \to \frac{\mathbb{E}[|X_1 - \lambda_0|^3]}{\lambda_0^{3/2}} < \infty\ \ \ \mathrm{as}\ \lambda_n \to \lambda_0. $$ follows from the fact that $$ \mathbb{E}[|X_1 - \lambda_n|^3] \to \mathbb{E}[|X_1 - \lambda_0|^3] < \infty,\\ \lambda_n^{3/2} \to \lambda_0^{3/2} > 0. $$
Therefore, from Berry-Esseen Theorem, we get $$ |F_n(x) - \Phi(x)| \leq \frac{C}{\sqrt{n}} \frac{\mathbb{E}[|X_1 - \lambda_n|^3]}{\lambda_n^{3/2}}, $$ where $F_n(x)$ is CDF of $\frac{\sqrt{n}(\bar{X}-\lambda_n)}{\sqrt{\lambda_n}}$, and $C$ is constant.
How can we transform the convergence(in law) "Qustion" from here?
Update in the end
The claimed convergence does not hold in general without assuming stronger convergence on $\lambda_n \to \lambda_0$.
In fact, for $\lambda_n = \lambda_0 - c\sqrt{\frac{\lambda_0}{n}}$, we can show $$ \frac{\sqrt{n}(\bar X - \lambda_n)}{\sqrt{\lambda_n}} \to N(c, 1), \text{ weakly}. $$
Too see this, we first note that the usual CLT for i.i.d. random variables implies $$ \frac{\sqrt{n}(\bar X - \lambda_0)}{\sqrt{\lambda_0}} \to N(0, 1), \text{ weakly}. $$
With this in mind, let us rewrite $\frac{\sqrt{n}(\bar X - \lambda_n)}{\sqrt{\lambda_n}}$ as $$ \frac{\sqrt{n}(\bar X - \lambda_n)}{\sqrt{\lambda_n}} = \frac{\sqrt{n}(\bar X - \lambda_0 + \lambda_0 - \lambda_n)}{\sqrt{\lambda_0}}\cdot \sqrt{\frac{\lambda_0}{\lambda_n}} = \frac{\sqrt{n}(\bar X - \lambda_0)}{\sqrt{\lambda_0}}\cdot \sqrt{\frac{\lambda_0}{\lambda_n}} + \frac{\sqrt{n}(\lambda_0 - \lambda_n)}{\sqrt{\lambda_n}} $$
The first term $\to N(0, 1)$ weakly as $n\to \infty$ since $\frac{\sqrt{n}(\bar X - \lambda_0)}{\sqrt{\lambda_0}} \to N(0, 1)$ and $\lambda_n \to \lambda_0$.
The second term $ = \frac{\sqrt{n}c\sqrt{\lambda_0/n}}{\sqrt{\lambda_n}} = \frac{c\sqrt{\lambda_0}}{\sqrt{\lambda_n}} \to c$ as $n\to \infty$.
Combining, we conclude that (first term) + (second term) $\to N(c, 1)$ weakly.
As you can see from above, the problem here is the convergence of $\lambda_n \to \lambda_0$ is too slow. By assuming faster convergence, we can show your claim.
Alternatively, you can prove $\frac{\sqrt{n}(\bar X - \lambda_0)}{\sqrt{\lambda_n}} \to N(0, 1)$ without assuming faster convergence of $\lambda_n$.
Update:
It seems like the author is assuming $X_1, \dots, X_n \sim \text{Pois}(\lambda_n)$ rather than $X_1, \dots, X_n \sim \text{Pois}(\lambda)$. The difference is that $\lambda_n$ changes as $n$ increases, so $X_1, \dots, X_n$ is not really a regular sequence, it is a triangular sequence ($X_i$ should be rather written as $X_{i,n}$ to explicitly show dependence on $n$.). In that case, all you have to do is to observe the right hand side of your inequality $$ |F_n(x) - \Phi(x)| \leq \frac{C}{\sqrt{n}} \frac{\mathbb{E}[|X_1 - \lambda_n|^3]}{\lambda_n^{3/2}}, $$ converges to zero as $n\to \infty$.