convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get 
$$\frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}}$$

A slightly better bound would be
$$\frac{2^n+3}{3^n-1}=\frac{2^n(1+3\cdot 2^{-n})}{3^n(1-3^{-n})}\leq \frac{5/2}{2/3}\frac{2^n}{3^n}$$
