<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] Notice,  $$\sum_{n=2}^{\infty}\frac{3}{10^n}$$
$$=3\sum_{n=2}^{\infty}\frac{1}{10^n}$$
$$=3\left(\underbrace{\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+\ldots}_{\text{sum of an infinite G.P.}}\right)$$
$$=3\left(\frac{\frac{1}{10^2}}{1-\frac{1}{10}}\right)$$
$$=3\left(\frac{1}{90}\right)=\color{red}{\frac 1{30}}$$
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