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

• Apr 25th 2010, 09:37 PM
alin1916
Determine whether the series is absolutely convergent, conditionally convergent ....
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

http://i40.tinypic.com/nl3c08.png
• Apr 25th 2010, 09:46 PM
Prove It
Quote:

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

http://i40.tinypic.com/nl3c08.png

Ratio test:

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

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

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

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

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

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

$= \frac{3}{2}$

$> 1$.

So the series is divergent.