The series converges, since |8/9| = 8/9 < 1.
Let [tex]S_n[/tex] denote the n-th partial sum of the series,
[tex]S_n=\displaystyle\sum_{k=0}^n\left(\frac89\right)^k[/tex]
[tex]S_n=1+\dfrac89+\left(\dfrac89\right)^2+\cdots+\left(\dfrac89\right)^n[/tex]
Multiply both sides by 8/9:
[tex]\dfrac89S_n=\dfrac89+\left(\dfrac89\right)^2+\left(\dfrac89\right)^3+\cdots+\left(\dfrac89\right)^{n+1}[/tex]
Subtract this from [tex]S_n[/tex] to eliminate all but the first term in
[tex]S_n-\dfrac89S_n=1-\left(\dfrac89\right)^{n+1}[/tex]
Solve for [tex]S_n[/tex]:
[tex]\dfrac19S_n=1-\left(\dfrac89\right)^{n+1}[/tex]
[tex]S_n=9-9\left(\dfrac89\right)^{n+1}[/tex]
As [tex]n\to\infty[/tex], the exponential term vanishes, so that
[tex]\displaystyle\lim_{n\to\infty}S_n=\sum_{k=0}^\infty\left(\frac89\right)^k=\boxed{9}[/tex]
