Final answer:
To determine whether the series is convergent or divergent, we can use the limit comparison test. By comparing the given series with a known convergent series, we find that the given series is convergent. Therefore, we conclude that the series is convergent based on the comparison test.
Step-by-step explanation:
To determine whether the series \sum\limits_{n=1}^{\infty} \frac{{5 - 7n}}{{n^3}} is convergent or divergent, we can use the limit comparison test. This test allows us to compare the given series with another series whose convergence or divergence is already known.
We will compare the given series with the series \sum\limits_{n=1}^{\infty} \frac{1}{{n^2}}. Let's find the limit of the ratio of corresponding terms:
\lim\limits_{n\to\infty} \frac{{\frac{{5 - 7n}}{{n^3}}}}{{\frac{1}{{n^2}}}} = \lim\limits_{n\to\infty} \frac{{5n^2 - 7n}}{{n^3}} = \lim\limits_{n\to\infty} \frac{{5 - \frac{7}{n}}}{{n^2}} = 0
Since the limit is finite and positive, it means that both series have the same behavior. Therefore, the given series is convergent.
The series in question is \(\sum_{n=1}^{\infty} \frac{5 - 7n}{n^3}\). To determine whether it is convergent or divergent, we can use the comparison test. As n approaches infinity, the term \(\frac{-7n}{n^3}\) will dominate the numerator, thus the series will behave similarly to \(-\sum_{n=1}^{\infty} \frac{7}{n^2}\).
Since \(\frac{7}{n^2}\) is a p-series with p=2 (which is greater than 1), it is known to be convergent. Following the comparison test, if a series \(a_n\) is always less than a convergent series \(b_n\) for all n, then \(a_n\) is also convergent. In this case, our series is negative but \(a_n\) in absolute value is less than the convergent p-series \(b_n = \frac{7}{n^2}\).
Therefore, we conclude that the series is convergent based on the comparison test.