Answer
$$9$$
Work Step by Step
A geometric series $\Sigma_{k=0}^{\infty} ar^k$ is said to be converges when $|r| \lt 1 $ and diverges when $|r| \geq 1$ and $\Sigma_{k=0}^{\infty} ar^k=\dfrac{a}{1-r}$
We have a series $\Sigma (\dfrac{9}{10})^k$ with $r=0.9 \lt 1$.
This means that the given series converges.
Now, $\Sigma_{k=1}^{\infty} (\dfrac{9}{10})^k=\dfrac{0.9}{1-0.9}=9$
