Is $(1+ z_n/n)^n - e^{z_n} $ always negative for $n \geq 2$ when $ \frac{z_n}{n} \rightarrow 0$?

Question

This question is similar to If $z_n \to z$ then $(1+z_n/n)^n \to e^z$

In this case $z_n$ does not converge but $ \frac{z_n}{n} \rightarrow 0 $.

Answer 1

Consider $$\ln\left[(1+z_n/n)^n\right] =n\ln(1+z_n/n)=z_n-\frac{z_n^2}{2n}+\cdots.$$ This will usually be less than $z_n$ eventually.