91Ó°ÊÓ

When discussing series, exponential growth is a pivotal concept. It often determines the behavior of a series, as it denotes how quickly its terms grow as we progress. In our case, the term \(\left( \frac{3}{2} \right)^n \) exhibits exponential growth. This means that as \(n\) becomes larger, each term of \(\left( \frac{3}{2} \right)^n\) grows larger at an exponential rate. Be mindful of series-looking components like these, for exponential growth quickly makes the series dominant, impacting convergence or divergence. Such growth can easily outweigh the polynomial decay represented by \(\frac{1}{n^3}\). Hence, this overpowering exponential factor is crucial in deciding the outcome of a series’ behavior.

Absolute convergence checks if the series \(\sum_{n=1}^{\infty} |a_n|\) is convergent. Simply put, it asks if ignoring the signs of terms still results in convergence. For our series \(\sum_{n=1}^{\infty} \frac{(-3)^{n}}{n^{3} 2^{n}}\), this involved checking \(\sum_{n=1}^{\infty} \frac{ (3/2)^{n}}{n^{3}}\). Unfortunately, despite the \(\frac{1}{n^3}\) part shrinking terms, the dominant \(\left(\frac{3}{2}\right)^n\) pulls it to divergence. Normally, a convergent absolute series guarantees the original series' convergence, but here no such assurance is provided due to explosive exponential growth.

Divergence occurs when a series does not converge to a finite sum. It's akin to an unsolvable puzzle due to certain aspects. In our compare-the-components game, \(\left( \frac{3}{2} \right)^n\) grows so fast that the smaller, shrinking \(\frac{1}{n^3}\) can't catch up, leading to the series' divergence. Though the series is alternating, indicating potential conditional convergence, the exponential factor undermines this, ensuring overall divergence. A divergent series defies taming by simply adding more terms, and shows increasing terms with no summed limit.

Series Comparison is a strategy involving measure or comparison against known behavior series. Our foundational task in this problem was to use the comparison test to determine convergence by likening \(\sum_{n=1}^{\infty} \frac{(3/2)^n}{n^3}\) to a known p-series, \(\sum_{n=1}^{\infty} \frac{1}{n^3}\). While a simple \(\frac{1}{n^3}\) converges, the exponential modifier \(\left(\frac{3}{2}\right)^n\) fostered faster growth, leading to divergence. Series Comparison helps us assess whether the series’ terms are larger or smaller than those from a known convergence or divergence reference, clarifying the nature of the series in question.