Limit of $i^{n!}-2^{-n}$

Question

I ran into this problem in Palka's Book which said to compute the limit of $z_n=i^{n!}+2^{-n}$. My approach was to consider the real and imaginary limits separately. Clearly the limit of the real part of the complex sequence is $0$. However, $$\mathrm{Im}(z_n)=\sin\left(\frac{\pi n!}{2}\right)$$ I don't think this converges. Appreciate any help. Thanks

Edit: On the other hand I just realized $i^{n!}$ is not always imaginary. Anyway hope someone could answer this

Answer 1

Hint for $n \geq 4$ we know that $n!$ is a multiple of $4$. Therefore, for $n \geq 4$ we have $i^{n!}=1$.

So your sequence becomes $$z_n=2^{-n}+1 \, \forall n \geq 4$$ Now, it is easy to consider this sequence, and the first three terms make absolutely no difference for the limit.

Answer 2

You are correct to say that $2^{-n}$ goes to 0. So instead we will focus on the imaginary part. We know that n is approaching infinity and we also know that 4 time any integer is still divisible by 4. Thus we can say n! is divisible by 4 and thus $i^{n!}$ approaches 1. So $z_n$ approaches 1.