(a) The n-th partial sum of the infinite series,
[tex]\displaystyle\sum_{n=0}^\infty\frac5{4^n}[/tex]
is
[tex]S_n = \displaystyle\sum_{k=0}^n\frac5{4^k} = 5\left(1+\frac14+\frac1{4^2}+\cdots+\frac1{4^n}\right)[/tex]
Multiplying both sides by 1/4 gives
[tex]\dfrac14S_n = \displaystyle\sum_{k=0}^n\frac5{4^k} = 5\left(\frac14+\frac1{4^2}+\frac1{4^3}+\cdots+\frac1{4^{n+1}}\right)[/tex]
Subtract this from [tex]S_n[/tex] and solve for [tex]S_n[/tex] :
[tex]S_n-\dfrac14S_n = 5\left(1-\dfrac1{4^{n+1}}\right)[/tex]
[tex]\dfrac34 S_n = 5\left(1-\dfrac1{4^{n+1}}\right)[/tex]
[tex]S_n = \dfrac{20}3\left(1-\dfrac1{4^{n+1}}\right)[/tex]
(your solution is also correct)
(b) The infinite sum is equal to the limit of the n-th partial sum:
[tex]\displaystyle\sum_{n=0}^\infty \frac5{4^n} = \lim_{n\to\infty} \boxed{\sum_{k=0}^n \frac5{4^k}}[/tex]
and the sum indeed converges to 20/3.
