# Can the Ratio & Root Tests show divergence directly?

An infinite series ⅀ $$a_n$$ is absolutely convergent if ⅀ $$|a_n|$$ is convergent. However, just because the absolute value of a series isn't convergent by some test doesn't mean it can't be conditionally convergent without the absolute value. This much I know.

So, when presented with series tests such as the Ratio and Root Tests, I'm a bit confused. They both have to do with the absolute value of a series. For example, the Ratio Test says:

Take $$L = \lim\limits_{n \rightarrow \infty} \left |\frac{a_{n+1}}{a_n} \right|$$

$$\bullet$$ If $$L < 1,$$ the the series converges absolutely.

$$\bullet$$ If $$L > 1,$$ the series diverges.

$$\bullet$$ If $$L = 1,$$ the test is inconclusive.

My question is, is this divergence absolute? For example, I know that if I put ⅀ $$(-1)^n \frac{n!}{9^n}$$ through the Ratio Test, I would end up with infinity, which is greater than zero. This would signal that the series diverges by the Ratio Test. Is that the end? Can I try some other test WITHOUT the absolute value that could signal conditional convergence?

If $$\lim_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}>1$$ or if $$\lim_{n\to\infty}\frac{\lvert a_{n+1}\rvert}{\lvert a_n\rvert}>1$$, then you don' have $$\lim_{n\to\infty}a_n=0$$. So, both series $$\displaystyle\sum_{n=0}^\infty\lvert a_n\rvert$$ and $$\displaystyle\sum_{n=0}^\infty a_n$$ diverge.