Answer
Diverges
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 $\Sigma 3(1.001)^k$ with $r=1.001 \gt 1$.
This means that the given series diverges.
