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?

You know that $\sum_{n\ge 1}q^n$ converges (you have $q=\frac{1}{10}$, similarily it works for all $q\in(-1,1)$), so let $$S=\sum_{n\ge 1}q^n=q+q^2+q^3+\dots$$ then $$qS=q^2+q^3+\dots = S-q$$ so$$S=\frac{q}{1-q}$$
Hence
$$\sum_{n=2}^{\infty}\frac{3}{10^n}=3\left(\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^n-\frac{1}{10}\right)=\frac{1}{30}$$
