$\text { The terms of a series } \sum _ { n = 1 } ^ { \infty } a _ { n } \text { are defined recursively. Determine the convergence or }$ divergence of the series. Explain your reasoning.
$a _ { 1 } = 6 , a _ { n + 1 } = \frac { 4 n - 1 } { - 3 n - 2 } a _ { n }$

Question

$\text { The terms of a series } \sum _ { n = 1 } ^ { \infty } a _ { n } \text { are defined recursively. Determine the convergence or }$ divergence of the series. Explain your reasoning. 
$a _ { 1 } = 6 , a _ { n + 1 } = \frac { 4 n - 1 } { - 3 n - 2 } a _ { n }$

Quizplus · Accepted Answer

The Ratio Test can be applied here by examining the limit of the absolute value of the ratio of consecutive terms, $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $. Given $ a_{n+1} = \frac{4n-1}{-3n-2}a_n $, the ratio $ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{4n-1}{-3n-2} \right| $ does not tend to a value less than 1 as $ n $ approaches infinity, indicating divergence.