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I would like to now, how to prove that this series diverges.

$$ 3-\frac{9}{2}+\frac{27}{4}-\frac{81}{8}+\frac{243}{16}... $$

How could I prove it? I have tried the n-term criterion that requires to prove that $\lim_{n \to \infty}a_n$ is not zero, but i dont know how to express $a_n$.

exist another method?

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    $\begingroup$ Hint: Geometric series and common ratios. $\endgroup$ – Simply Beautiful Art Mar 5 '17 at 22:43
  • $\begingroup$ Check the sums $n_2 + n_3$, $n_4 + n_5,$ $\dots$. $\endgroup$ – Ninja Mar 5 '17 at 22:45
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    $\begingroup$ The $n$-th term seems to be $3 \cdot (-3/2)^n$,so you have a classic geometric series. Since $|-3/2|>1$, the series is divergent $\endgroup$ – Crostul Mar 5 '17 at 22:45
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You can expres the series how follows

$$ 3-\frac{9}{2}+\frac{27}{4}-\frac{81}{8}+\frac{243}{16}... = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^n}{2^{n-1}} $$ and $$ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^n}{2^{n-1}}=\sum_{n=1}^{\infty} -1(-1)^{n} \frac{3^n}{2^{-1}2^{n}}= \sum_{n=1}^{\infty} \frac{-1}{2^{-1}}(-1)^{n} \frac{3^n}{2^{n}} $$ agrouping the equal powers it gives:

$$ = \sum_{n=1}^{\infty} -2 \left(-\frac{3}{2}\right)^n $$

Then, this is a geometric series and diverges by the ratio convergence criterion $|-3/2| >1$

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