<s>[INST] evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$ 
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$

I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how do I find the sum?
 [/INST] $$\sum^{\infty}_{n=2} \frac{3}{10^n}=\frac{3}{100}+\frac{3}{1000}+...=0.03+0.003+...=0.0333...=\frac{0.333...}{10}=\frac{\frac{1}{3}}{10}=\frac{1}{30}$$
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