sequence /series convergence of 2^2n 3^(1-n)

this step in the proof is confusing me:

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

cheers,

gregg

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You can use $ signs, like $x^2+y^2$ to give$x^2 + y^2$around the latex. Also infinity is \infty:$\infty$. Your question isn't very clear. Please format is using the $ signs. –  Aryabhata Dec 3 '10 at 7:10
Thanks for the tips Moron, I hope that's readable now? –  gnicezw Dec 3 '10 at 7:27
Yes, it is readable now. As to your question, write $4^n = 4 \times 4^{n-1}$. –  Aryabhata Dec 3 '10 at 7:28
Moron, Jonas thanks for clearing that up - newbie to the site and calculus so I appreciate your help. –  gnicezw Dec 3 '10 at 7:37

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:
$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$