# Complex Sequence Limits

• Aug 13th 2010, 07:14 AM
emathinstruction
Complex Sequence Limits
I am asked to prove that if $\displaystyle z\in\mathbb{C}$ then $\displaystyle \displaystyle lim_{n\rightarrow\infty} \frac{z^n}{n^n}=0$ and $\displaystyle \displaystyle lim_{n\rightarrow\infty} \frac{z^n}{n!}=0$

So I assume I have to treat these as sequence because $\displaystyle \frac{z^n}{n^n}$ cannot be a function of z since the n is there. I know the definition of the limit but I just get stuck. In $\displaystyle z$ were real I think I would take the log and proceed from there, but that seems considerably harder with the complex logarithm.

Anyway I have:

Let $\displaystyle \epsilon >0$. Now I have to find a $\displaystyle K$ such that when $\displaystyle n>K$, $\displaystyle |\frac{z^n}{n^n}-0|<\epsilon$ which I cannot do.

Can anyone help me... thanks.
• Aug 13th 2010, 07:22 AM
InvisibleMan
Well if you want you can use the fact that
z= r exp(it) where r and t are real and then look on the real sequence of $\displaystyle |z|^n/n^n=r^n/n^n$, where presumably you have proven it already in your real analysis course.