$$S_n=\frac{2^{n+2}-4}{2^n}$$

$$f_n=S_n-S_{n-1}$$

$$f_n=\frac{2^{n+2}-4}{2^n}-\frac{2^{n+1}-4}{2^{n-1}}$$

$$f_n=\frac{2^{n+2}-4}{2^n}-\frac{2^{n+1}-4}{2^n·2^{-1}}$$

$$f_n=\frac{2^{n+2}-4}{2^n}-\frac{\left(2^{n+1}-4\right)·2^1}{2^n}$$

$$f_n=\frac{\cancel{2^{n+2}}-4\cancel{-2^{n+2}}+8}{2^n}$$

$$f_n=\frac{-4+8}{2^n}$$

$$f=\underset{n\to \infty }{lim}{f}_n=\frac{4}{2^n}=\frac{4}{2^{\infty }}=\frac{4}{\infty }$$

$$f=0$$

Let's examine the series through its members, is it a geometric series?

$$f_1=\frac{4}{2^1}=2$$

$$f_2=\frac{4}{2^2}=\frac{4}{4}=1$$

$$f_3=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}$$

$$f_4=\frac{4}{2^4}=\frac{4}{16}=\frac{1}{4}$$

$$q=\frac{f_2}{f_1}=\frac{f_3}{f_2}=\frac{f_4}{f_3}=...=\frac{1}{2}$$

$$S=\underset{n\to \infty }{lim}{S}_n=\frac{f_1}{1-q}, \left|q\right|<1$$

$$S=\frac{2}{1-{\displaystyle \frac{1}{2}}}=\frac{2}{{\displaystyle \frac{1}{2}}}$$