Find all points of accumulation of $(a_n)$ in $\mathbb C$, whereby $$a_n= \left(1+\frac{i^{2n}}{n}+\frac{1}{2n}\right)^n+\frac{ne^{-in}}{\exp(\log(n)^2)}.$$
My thoughts:
I first looked at the second term, and found that $$\exp(\log(n)^2)=\exp(2\log(n))=n^2,$$ and thereby $$\frac{ne^{-in}}{\exp(\log(n)^2)}=\frac{1}{ne^{in}},$$ which tends to zero so it can be ignored (Is this correct?).
For the first term I wrote, $$\left(1+\frac{i^{2n}}{n}+\frac{1}{2n}\right)^n=\left(1+\frac{(-1)^n}{n}+\frac{1}{2n}\right)^n$$
Looking at $$a_{2n}=\left(1+\frac{1}{2n}+\frac{1}{2 × 2n}\right)^{2n}$$ and at $$a_{2n-1}=\left(1+\frac{-1}{2n-1}+\frac{1}{2 × (2n-1)}\right)^{2n-1},$$ I get stumped. Any help?
That's correct. Now for finding the limits of $a_{2n}$ and $a_{2n+1}$ we have: $$a_{2n}=(1+\dfrac{\frac{3}{2}}{2n})^{2n}=(1+\dfrac{\frac{3}{2}}{2n})^{{\dfrac{2n}{3\over 2}}\dfrac{3}{2}}\to e^\dfrac{3}{2}$$and $$a_{2n+1}=(1-\dfrac{1}{2(2n+1)})^{2n+1}\to e^{-\dfrac{1}{2}}$$so the accumulation points are $(e\sqrt e,0)$ and $(\dfrac{1}{\sqrt e},0)$