Find the expected value of $\sqrt{K}$ where $K$ is a random variable according to Poisson distribution with parameter $\lambda$.
I don't know how to calculate the following sum:
$E[\sqrt{K}]= e^{-\lambda} \sum_{k=0}^{\infty} \sqrt{k} \frac{\lambda^k}{k!} $
Based on Wiki (https://en.wikipedia.org/wiki/Poisson_distribution) I know that should be approximately $\sqrt{\lambda}$.
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 $$E(\sqrt {P_\lambda})=\sqrt \lambda(1-\frac 1 8 \lambda^{-1}+o(\lambda^{-{3/2}})).$$