# 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

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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

## 1 Answer

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:

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

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Thanks again mate, simple but had me stumped. – gnicezw Dec 3 '10 at 7:42
@gnicezw: You can show your appreciation by voting the answer up (the arrow near the top left of the answer). You can even accept the answer by clicking on the tick mark. btw, Welcome to this site! – Aryabhata Dec 3 '10 at 8:05
@gnicezw: Welcome to the forum. The best thing is to press the accept, otherwise this question will always be listed as "open". – AD. Dec 3 '10 at 8:26
(Users with less than 15 points can't vote, and one answer with positive vote total suffices to remove a question from the Unanswered list.) – Jonas Meyer Dec 3 '10 at 16:47