Is $\lim_{n\rightarrow\infty }nz^{n!n}=0$ for $|z|<1$?

Question

Is $\lim_{n\rightarrow\infty }nz^{n!n}=0$ for $|z|<1$? We have a $\infty \cdot 0$ case, then how we proceed? How to use the L'Hospital's Rule?

Thanks in advance!

Answer 1

It's enough to prove it for $0<z<1$. Define $a_n:=n z^{n! n}$. Then $\frac{a_{n+1}}{a_n}=\frac{n+1}nz^{(n+1)!(n+1)-n!n}$. Since $(n+1)!(n+1)-n!n\geqslant (n+1)n!n-n!n=n!n²$, $\frac{a_{n+1}}{a_n}=\frac{n+1}nz^{n^2\cdot n!}\leqslant 2z^{n^2\cdot n!}\lt 1/2$ for $n$ large enough.

Answer 2

Without any complex math, consider what happens to the module of this number: $n|z|^{n!n}<<<n|z|^n\rightarrow 0$ if $|z|<1$.