# Thread: Determine whether the series is absolutely convergent, conditionally convergent ....

1. ## Determine whether the series is absolutely convergent, conditionally convergent ....

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

2. Originally Posted by alin1916
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Ratio test:

$\displaystyle \lim_{k \to \infty}\left|\frac{a_{k + 1}}{a_k}\right|$

$\displaystyle = \lim_{k \to \infty}\left|\frac{\frac{(-1)^{k + 2}\cdot 1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2) \cdot (3k + 1)}{(k + 1)!2^{k + 1}}}{\frac{(-1)^{k + 1}\cdot 1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2)}{k!2^k}}\right|$

$\displaystyle = \lim_{k \to \infty}\frac{\frac{1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2) \cdot (3k + 1)}{(k + 1)!2^{k + 1}}}{\frac{1 \cdot 4 \cdot 7 \cdot \dots \cdot (3k - 2)}{k!2^k}}$

$\displaystyle = \lim_{k \to \infty}\frac{3k + 1}{2(k + 1)}$

$\displaystyle = \lim_{k \to \infty}\frac{3k + 1}{2k + 2}$

$\displaystyle = \lim_{k \to \infty}\frac{3}{2}$ by L'Hospital's Rule

$\displaystyle = \frac{3}{2}$

$\displaystyle > 1$.

So the series is divergent.