Limit of $a_n:=\left(\frac{i}{2}\right)^n$

Question

Study the convergence and find the sequences limit $$a_n:=\left(\frac{i}{2}\right)^n.$$

\begin{align} \lim_{n\to\infty}\left(\frac{i}{2}\right)^n & =\lim_{n\to\infty}\left(\frac{1}{2}\left(\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\right)\right)^n \\ & =\lim_{n\to\infty}\left(\frac{1}{2^n}\left(\cos\left(\frac{n\pi}{2}\right)+i\sin\left(\frac{n\pi}{2}\right)\right)\right)=0. \end{align}

Is my answer right? Thanks in advance!

Answer 1

That the last limit is equal to $0$ is no more ovious (to me, at least) than the first one.

You can just say that$$\lim_{n\to\infty}\left\lvert\left(\frac i2\right)^n\right\rvert=\lim_{n\to\infty}\left\lvert\frac i2\right\rvert^n=\lim_{n\to\infty}\frac1{2^n}=0$$and that therefore$$\lim_{n\to\infty}\left(\frac i2\right)^n=0.$$ So, there is no need to use $\pi$ or trigonometric functions.

Answer 2

Starting from $n=0$, the sequence of numerators is periodically

$$1,i,-1,-i,1,i,-1,-i,\cdots$$

Hence you can consider four intertwined sequences

$$\frac1{2^{4n}},\frac i{2^{4n+1}},-\frac1{2^{4n+2}},-\frac i{2^{4n+3}},$$

which all converge to $0$.