How to prove an inequality $\left| {g(j + 1)} \right| \le 5/4$ in Stein's method for Poisson approximation

Question

The following is a lemma in

Barbour, A. D., Holst, L., & Janson, S. (1992). Poisson approximation. Oxford: Clarendon Press,p7.

For $j=1,2,...$ and $\lambda > 0$, we have

$\left| {g(j + 1)} \right| \le {(\lambda - j)^{ - 1}}$ if $j < \lambda $;

and

$\left| {g(j + 1)} \right| \le \frac{{j + 2}}{{(j + 1)(j + 2 - \lambda )}}$ if $j > \lambda - 2$.

How to prove that $\left| {g(j + 1)} \right| \le 5/4$ for all $\lambda > 0$ and $j=1,2,...$?

The book ignores the proof.

Answer 1

Let $k = \frac{(2j + 1)(j + 2)}{2j + 3}$. It's easy to see that $k$ is strictly between $j$ and $j + 2$. If $\lambda \ge k$, then by the first inequality, $|g(j + 1)| \le \frac{1}{\lambda - j} \le \frac{1}{k - j} = \frac{2j + 3}{2j + 2}$. If $\lambda \le k$, then by the second inequality, $|g(j + 1)| \le \frac{(j + 2)}{(j + 1)(j + 2 - \lambda)} \le \frac{(j + 2)}{(j + 1)(j + 2 - k)} = \frac{2j + 3}{2j + 2}$. So either way, $|g(j + 1)| \le \frac{2j + 3}{2j + 2}$. Because $j \ge 1$, it follows that $|g(j + 1)| \le \frac{5}{4}$.