$$S_n=\frac{5^{n+1}-5}{4·5^n}$$

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

$$f_n=\frac{5^{n+1}-5}{4·5^n}-\frac{5^n-5}{4·5^{n-1}}$$

$$f_n=\frac{5^{n+1}-5}{4·5^n}-\frac{5^n-5}{4·5^n·5^{-1}}$$

$$f_n=\frac{5^{n+1}-5}{4·5^n}-\frac{\left(5^n-5\right)·5^1}{4·5^n}$$

$$f_n=\frac{5^{n+1}-5}{4·5^n}-\frac{5^{n+1}-5·5}{4·5^n}$$

$$f_n=\frac{\cancel{5^{n+1}}-5\cancel{-5^{n+1}}+5^2}{4·5^n}$$

$$f_n=\frac{-5+25}{4·5^n}$$

$$f_n=\frac{20}{4·5^n}$$

$$f_n=\frac{4·5}{4·5^n}$$

$$f=\underset{n\to \infty }{lim}{f}_n=\frac{1}{5^{n-1}}=\frac{1}{5^{\infty }}=\frac{1}{\infty }$$

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

$$f_1=\frac{1}{5^{1-1}}=\frac{1}{5^0}=\frac{1}{1}=1$$

$$f_2=\frac{1}{5^{2-1}}=\frac{1}{5^1}=\frac{1}{5}$$

$$f_3=\frac{1}{5^{3-1}}=\frac{1}{5^2}$$

$$f_4=\frac{1}{5^{4-1}}=\frac{1}{5^3}$$

$$q=\frac{f_2}{f_1}=\frac{f_3}{f_2}=\frac{f_4}{f_3}=...=\frac{1}{5}$$

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

$$S=\frac{1}{1-{\displaystyle \frac{1}{5}}}=\frac{1}{{\displaystyle \frac{4}{5}}}$$