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This is super elementary, so I apologize in advanced:

$$\frac{4}{3} - \frac{8}{9} + \frac{16}{27} - \frac{32}{81} + \ldots$$

Does this series converge or diverge?

I am trying to find a pattern but neither taking out $4/3$ or $4$ seems to work any suggestions would be greatly appreciated. I am sorry again for being so elementary. I am not in this class just trying to refresh myself with algebra for tutoring

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    $\begingroup$ Is that it or is it supposed to have more than 4 terms? $\endgroup$ – user228113 Apr 30 '15 at 20:28
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    $\begingroup$ Do you mean $$\sum_{n=2}^{\infty }\frac{(-2)^n}{3^{n-1}}$$ $\endgroup$ – E.H.E Apr 30 '15 at 20:29
  • $\begingroup$ Try writing the denominators as powers of 3 and the numerators as powers of 2. The pattern will become super obvious. $\endgroup$ – John Joy Apr 30 '15 at 20:36
  • $\begingroup$ Obvious now, thanks and sorry for the trivial post $\endgroup$ – Wolfy Apr 30 '15 at 20:44
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The series is converging.

As you can see, the terms are given by $x_n=(-1)^{n-1}\frac{2^{n+1}}{3^n}=(-1)^{n-1}\cdot2\cdot (\frac{2}{3})^n$ where $n\geq1$

Now, we have

$|x_{n+1}|-|x_n|=2(\frac{2}{3})^{n+1}-2(\frac{2}{3})^n<0$

$\lim_{n\to\infty} |x_n|=\lim_{n\to\infty}2\cdot (\frac{2}{3})^n=0$ since $\frac{2}{3}<1$

Thus, by the alternating series test, the series converges.

For the sake of completion, put $\sum x_n=\sum_{n=1}^\infty -2\cdot(\frac{-2}{3})^n =-2\sum_{n=1}^\infty (\frac{-2}{3})^n=-2(-\frac{2}{5})=\frac{4}{5}$

QED.

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