To prove that $\lim_{n\to \infty}nz^{n+1} = 0$

Question

I came across with the following limit in a complex analysis course and I'm not sure how to prove it by definition. $\lim_{n\to\infty} n z^{n+1}$ where $|z|<1$

Answer 1

We prove that $\lim_{n\to \infty}nz^{n+1} = 0$. Let's rewrite $z$ in a ploar form $z = r (\cos \theta + i \sin \theta)$, where $r = |z| < 1$. Then, $ |nz^{n+1} - 0| = |nz^{n+1}| = n r^{n+1} |\cos \theta + i \sin \theta|^{n+1} = n r^{n+1}. $

Now, we rewrite $r = \frac{1}{R + 1}$ for some $R>0$. Note that $r^{n+1} = \frac{1}{(R + 1)^{n+1}} < \frac{1}{nR + \frac{(n-1)n}{2}R^2}$, since $(R + 1)^{n+1} = \sum_{i=0}^{n+1}\binom{n}{i}R^i > \binom{n}{1}R + \binom{n}{2}R^2$.

Thus, $|nz^{n+1} - 0| < \frac{1}{R + \frac{(n-1)}{2}R^2} < \epsilon$ for $\frac{2}{R^2}(\frac{1}{\epsilon} - 1) + 1 < n_0$.