Why this :$ \lim_{n\to \infty } n (\frac{1+i}{2})^n=0$?

Question

I have tried to show that :$ \lim_{n\to \infty } n (\frac{1+i}{2})^n=0$ with $i$ is unit imaginary part , I only guess that is 0 because I have $0<|\frac{1+i}{2}|<1$ then limit should be $0$ , but i'm note care about that !!!

Answer 1

Note that if $a_n$ is a sequence of complex numbers $\lim \ a_n = 0 $ iff $\lim \ |a_n| = 0 $ where |.| is the modulus.

$|\frac{1+i}2| = \sqrt{ \frac 14 + \frac 14} = \frac 1 {\sqrt 2} <1 $

$|\frac{1+i}2| = \frac 1 {\sqrt 2} e^{\frac{i\pi}4 }$

$|na_n| = n(\frac 1 {\sqrt 2})^n $ which converge to $0$. Indeed, $ \forall r<1, \ \lim(n r^n) = 0 $

Therefore,

$ \lim \ na_n = 0$

Answer 2

We have

$$\frac{1+i}{2}=\frac1{\sqrt 2}e^{i\frac{\pi}4}$$

and then

$$\left(\frac{1+i}{2}\right)^n=\frac1{\sqrt {2^n}}e^{i\frac{n\pi}4}$$

and thus

$$0\le \left|n \left(\frac{1+i}{2}\right)^n\right|=\frac{n}{\sqrt {2^n}}\to0$$

Answer 3

Hint:$ \lvert n\cdot (\frac {1+i}2)^n\rvert =\frac {{\sqrt2}^nn}{2^n}=\frac n{2^{\frac n2}}$, and $n=\mathcal o(\sqrt2^n)$...