I am once again stuck on some asymptotics.
In this proof, it is assumed that $1/2\le \lambda \le n/2$.
First, it is shown that $$E(Y) \le ne^{-\lambda}e^{(\lambda/n)}.$$
Next, it is shown that $$ne^{-\lambda}e^{(-\lambda^2/n)} \le E(Y).$$
The conclusion is that $E(Y) = ne^{-\lambda} (1 + O(\lambda^2/n)).$
I don't understand why the lower bound is necessary here. Isn't it true that in this case, $e^{\lambda/n} = 1 + O(\lambda/n)$ because $$e^{\lambda/n} = 1 + \frac{{(\lambda/n)}^2}{2!} + \frac{{(\lambda/n)}^3}{3!} + \cdots \le 1 + \lambda/n \left( 1 + \frac{1}{2!} + \frac{1}{3}! + \cdots \right) = 1+e \frac{\lambda}{n} = 1+ O(\frac{\lambda}{n})?$$
In the inequality, I used the fact that $(\frac{\lambda}{n})^k \le \frac{\lambda}{n}$ since $\lambda \le \frac{n}{2}$.
Am I missing something here? Why is the lower bound needed?
The upper bound shows that there is a function $g(n,\lambda) = O(\lambda/n)$ such that
$$ (ne^{-\lambda})^{-1} E(Y) - 1 \leq g(n,\lambda). \tag{1} $$
However, in order to conclude that $(ne^{-\lambda})^{-1} E(Y) - 1 = O(\lambda/n)$, you need to know that
$$ \left|(ne^{-\lambda})^{-1} E(Y) - 1\right| \leq C\lambda/n \tag{2} $$
for some constant $C$.
What if $(ne^{-\lambda})^{-1} E(Y) - 1 = -\lambda n$? Then we could just take $g(n,\lambda) = 0$, so $(1)$ would be true, but $(2)$ certainly wouldn't be.
The point of also showing that $ne^{-\lambda}e^{(-\lambda^2/n)} \le E(Y)$ is to find another function $h(n,\lambda) = O(\lambda/n)$ such that
$$ (ne^{-\lambda})^{-1} E(Y) - 1 \geq h(n,\lambda). \tag{3} $$
Then, $(1)$ and $(3)$ together imply $(2)$, which is the desired result, because
$$ \left|(ne^{-\lambda})^{-1} E(Y) - 1\right| \leq \max\{|g(n,\lambda)|,|h(n,\lambda)|\} = O(\lambda/n). $$