Let $x\sim\text{Pois}(\lambda)$, prove that for any $\lambda>0$, $$f(\lambda)=\text{E}(\log(x+0.5)-\log\lambda)+0.02/\lambda>0.$$ In other words, $$\sum_{x=0}^{\infty}\log(x+0.5)\frac{\lambda^x}{x!}e^{-\lambda}-\log\lambda+0.02/\lambda>0.$$ Here, $\log$ is the natural logarithm.
A possible approach: we have $f^{'}(\lambda)=\text{E}(\log(\frac{x+1.5}{x+0.5}))-1/\lambda-0.02/\lambda^2$ and $\lim_{\lambda\to\infty}f(\lambda)=0$. Please try to avoid large-scale numerical computations as much as possible, because analyzing the error in floating-point arithmetic of computational code can be troublesome.