Find $\lim_{n \to \infty} n(\frac{1+i}{2})^n$ Find $\lim_{n \to \infty} n(\frac{1+i}{2})^n$.

I don't know how to solve this limit. Should I use the fact that $\lim_{n \to \infty} n(\sqrt{2}/2)^n\cos(n \pi / 4)$ and $\lim_{n \to \infty} n(\sqrt{2}/2)^n\sin(n \pi / 4)$ for the real et imaginary part of $n(\frac{1+i}{2})^n$.

Can anyone give me a hint to solve the problem?
One may use the fact that, for any $|q|&lt;1$, we have
$$
\lim_{n \to \infty}n|q|^n=0.
$$ Apply it here, observing that
$$
\left|n\left(\frac{1+i}{2}\right)^n\right|\leq n\left|\frac1{\sqrt{2}}\right|^n.
$$
Consider that :

$$(1+i)^4 = ((1+i)^2)^2=(1+2i-1)^2=4i^2=-4$$

So for any $n=4(n\,div\,4)+(n\,mod\,4)$ we have :

$$\left|n\left(\frac{1+i}{2}\right)^n\right|=n2^{-n+2(n\,div\,4)+(n\,mod\,4)}|(1+i)^{n\,mod\,4}|$$

Now the magnitude of this cannot ever be $&gt;1$ so this converges to zero ( converging along distinct sets of values, but converging all the same to zero ).