# Does this alternating series diverge: $\sum_{i=1}^{\infty} \frac {(-1)^{i+1}\cdot 1\cdot 4 \cdot 7 \cdots (3i-2)}{i!2^i}$?

$$\sum_{i=1}^{\infty} \frac {(-1)^{i+1}\cdot 1\cdot 4 \cdot 7 \cdots (3i-2)}{i!2^i}$$

By the alternating series test and the ratio test, I found that this series does not absolutely converge. However, I'm not at all sure how to figure out whether it converges conditionally or diverges.

## 2 Answers

HINT Note that $$\dfrac{1\cdot 4 \cdot 7\cdots (3i-2)}{i! 2^i} \geq \dfrac{3\cdot 6 \cdots (3i-3)}{i! 2^i} = \dfrac{3^{i-1}(i-1)!}{i!2^i} = \dfrac1{3i} \left(\dfrac32\right)^i$$

Since $\displaystyle\lim_{i\to\infty}\frac{|a_{i+1}|}{|a_i|}=\lim_{i\to\infty}\frac{1\cdot4\cdot7\cdots(3i-2)(3i+1)}{(i+1)!2^{i+1}}\cdot\frac{i!2^i}{1\cdot4\cdot7\cdots(3i-2)}$

$\;\;\;$$=\displaystyle\lim_{i\to\infty}\frac{3i+1}{2(i+1)}=\frac{3}{2}>1, the series \displaystyle\sum_{i=1}^{\infty}|a_i| diverges by the Ratio Test and therefore \;\;\;\;\;$$\displaystyle\sum_{i=1}^{\infty}a_i$ diverges also since $\displaystyle\lim_{i\to\infty}a_i\ne0$.