Limit of $\lim_{n\to \infty}\frac{1}{n^{1/2}}\sum_{k=0}^n(\frac{1}{2^{1/2}}(1+i))^k$

Question

$\lim_{n\to \infty}\frac{1}{n^{1/2}}\sum_{k=0}^n(\frac{1}{2^{1/2}}(1+i))^k$

Does that series converge or diverge? I know that the limit of $\frac{1}{n^{1/2}}$ is $0$ and $|1+i|^k \le2^k$ and $\frac{1}{2^{1/2}}$ shouldn't really matter. I guess the result is $0$ because $0 *2^k=0$ for every k. Is that correct?

Answer 1

Note that $\zeta=(1+i)/\sqrt2=\exp(\pi i/4)$. Therefore $$S_n=\sum_{k=0}^n\zeta^k=\frac{\zeta^{n+1}-1}{\zeta-1} =\frac{\exp(\pi i(n+1)/4)}{\zeta-1}.$$ As $|\zeta^{n+1}|=1$, this sequence is bounded, so that $n^{-1/2}S_n\to0$.