# How to prove that this series diverges: $3-\frac{9}{2}+\frac{27}{4}-\frac{81}{8}+\frac{243}{16}+\dots$

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 don't know how to express $$a_n$$.

exist another method?

• Hint: Geometric series and common ratios. – Simply Beautiful Art Mar 5 '17 at 22:43
• Check the sums $n_2 + n_3$, $n_4 + n_5,$ $\dots$. – Ninja Mar 5 '17 at 22:45
• 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 – Crostul Mar 5 '17 at 22:45

$$3-\frac{9}{2}+\frac{27}{4}-\frac{81}{8}+\frac{243}{16}... = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^n}{2^{n-1}}$$ 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 $|-3/2| >1$