Expected Value of Square Root of Poisson Random Variable

In general, for smooth $g(X)$ you can do a Taylor expansion around the mean $\mu=E(X)$:

$$g(X)=g(\mu) + g'(\mu)(X-\mu)+ \frac{g^{''}(\mu)}{2!}(X-\mu)^2+ \frac{g^{'''}(\mu)}{3!}(X-\mu)^3+\cdots$$

So

$$E[g(X)]=g(\mu) + \frac{g^{''}(\mu)}{2!}m_2+ \frac{g^{'''}(\mu)}{3!} m_3+\cdots $$

where $m_i$ is the $i$-th centered moment. In our case $m_2=m_3 =\lambda$, so:

$$E[g(X)]=\sqrt{\lambda} - \frac{\lambda^{-1/2}}{8} + \frac{ \lambda^{-3/2}}{16} +\cdots $$

This approximation is useful only if $\lambda \gg 1$

Around $\lambda=0$ you have already written the series expansion so $$E(\sqrt{ P_\lambda})=(1-\lambda+O(\lambda^2))(\lambda+\frac {\sqrt 2}2\lambda^2+O(\lambda^3))=\lambda-(1-\frac {\sqrt 2}2)\lambda^2+O(\lambda^3)$$ Around $\lambda=\infty$, $\epsilon_\lambda=\frac {P_\lambda-\lambda}\lambda$ is tightly concentrated (approximately normal) around $0$ with mean $0$ and variance $\frac 1 \lambda$, so you can expand $\sqrt {P_\lambda}=\sqrt \lambda(1+\frac 1 2\epsilon_\lambda-\frac 1 8\epsilon_\lambda^2+O(\epsilon_\lambda^3))$ so that $$E(\sqrt {P_\lambda})=\sqrt \lambda(1-\frac 1 8 \lambda^{-1}+o(\lambda^{-{3/2}})).$$