$$S_n=\frac{2n+i^n}{n}$$ $$ \alpha= \lim_{n\to \infty }S_n = 2 $$
$$\Vert S_n-\alpha \Vert \lt \epsilon $$ I substitute into the equation $$\Vert S_n-\alpha \Vert \lt \epsilon $$ To get $$\Vert \frac{2n+i^n}{n}-2 \Vert \lt \epsilon $$ I simplified to get $$\Vert \frac{i^n}{n} \Vert \lt \epsilon $$ Then $$\Vert \sqrt[n]\frac{i^n}{n} \Vert \lt \sqrt[n]\epsilon $$ $$\Vert \frac{i}{\sqrt[n]n} \Vert \lt \sqrt[n]\epsilon $$ $$\frac{1}{\sqrt[n]n} \lt \sqrt[n]\epsilon $$ Rearranged the equation $$\frac{1}{\sqrt[n]\epsilon} \lt {\sqrt[n]n} $$ $$(\frac{1}{\sqrt[n]\epsilon})^n \lt ({\sqrt[n]n})^n $$ $$\frac{1^n}{\epsilon} \lt {n} $$ Finally $$\frac{1}{\epsilon} \lt \frac{n}{1^n} $$ So this is the answer i got. Help would be appreciated.