sequence /series convergence of 2^2n 3^(1-n) this step in the proof is confusing me:

$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$

please explain how/why this happened?

cheers,

gregg
The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator.  Maybe seeing a couple of extra steps will help:

$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$
4^n is really 4 * 4 * 4 * 4 …. 4^n
so by taking one of those 4's out of 4^n, it becomes 4^n-1