Convergence of $a_n = (1+i)^n+(1-i)^n$

Question

I am asking a question related to Is there a formula for $(1+i)^n+(1-i)^n$?

I am looking on the exact same term, just as a sequence, so i want to find out:

Is $a_n = (1+i)^n+(1-i)^n$ convergent or not?

My attempt is to use the binomial theorem: $$a_n = (1+i)^n+(1-i)^n=$$

$$\sum_{k=0}^\infty\binom{n}{k}i^k+\sum_{k=0}^\infty\binom{n}{k}(-i)^k$$ So i is either $1,-1,i,$ or $ -i$. What else do we know. I know about eulers formula, De Moivre's formula, from the link above i know that $$a_n = (1+i)^n+(1-i)^n$$ is twice the real part of $(1+i)^n$ (?), I also know that $(x+iy)(x-iy) = x^2+y^2$. Hm. Another thought i had: For every n we get a complex number, that has a real part x and a imaginary part y. I might use that $|z| = \sqrt{x^2+y^2}$ to define a majorant for my sequence?

I am pretty new to complex numbers, if you could help me on this that would be cool.

Answer 1

Note that $(1+i)(1-i) = 1^2 - i^2 = 1 + 1 = 2$. So your series diverges badly.