How to find the value of $\sum _{k=1}^{\mathrm{\infty}}(\frac{1}{9}{)}^{k}$ using partial sums?

So I was trying to prove an infinite sum by looking at the partial sum, when I ran into a problem.

Consider:

$\sum _{k=1}^{n}{\left(\frac{1}{9}\right)}^{k}=\frac{1}{8}{9}^{-n}({9}^{n}-1)$

but as there are no solutions to ${9}^{-n}({9}^{n}-1)=1$, is it possible to prove that

$\sum _{k=1}^{\mathrm{\infty}}{\left(\frac{1}{9}\right)}^{k}=\frac{1}{8}$

by the partial sum?

So I was trying to prove an infinite sum by looking at the partial sum, when I ran into a problem.

Consider:

$\sum _{k=1}^{n}{\left(\frac{1}{9}\right)}^{k}=\frac{1}{8}{9}^{-n}({9}^{n}-1)$

but as there are no solutions to ${9}^{-n}({9}^{n}-1)=1$, is it possible to prove that

$\sum _{k=1}^{\mathrm{\infty}}{\left(\frac{1}{9}\right)}^{k}=\frac{1}{8}$

by the partial sum?