Dividing each term in the given series by 0.315 reveals a simple geometric sum,
0.315 (1 + 1/1,000 + 1/1,000,000 + …)
or
0.315 (1 + 1/10³ + 1/10⁶ + …)
or
[tex]\displaystyle 0.315 \sum_{n=0}^\infty \frac1{10^{3n}} = 0.315 \sum_{n=0}^\infty \frac1{1000^n}[/tex]
i.e. a geometric sum with a common ratio of 1/1,000. I'm not sure what your instructor expects exactly, but you may already know that
[tex]\displaystyle a\sum_{n=0}^\infty r^n = \frac a{1-r}[/tex]
if |r | < 1. This is the case here, so
0.315 (1 + 1/1,000 + 1/1,000,000 + …) = 0.315 / (1 - 1/1,000)
… = (315/1000) / (999/1000)
… = 315/999
… = 35/111
