Question

In Exercises \(59-62,\) verify that the Ratio Test is inconclusive for the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}} $$

Short answer

By applying the Ratio Test to the provided p-series, the limit of the absolute ratio of the (n+1)th term to the nth term, as n approaches infinity, is found to be 1. Therefore, the Ratio Test is inconclusive for this series.

Step-by-step solution

Identify the p-series

The given series is \(\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}}\). In this series, 'p' is \(3 / 2\).

Apply the Ratio Test

The Ratio Test requires to calculate the limit as n approaches infinity of the absolute ratio of the (n+1)th term and nth term of the series. So, calculate the limit as n goes to infinity of \(\left|\frac{\frac{1}{(n+1)^{3/2}})}{\frac{1}{n^{3/2}}}\right|\). This simplifies to \(\lim_{n \to \infty} \left| \frac{n^{3/2}}{(n+1)^{3/2}} \right|\).

Evaluate the limit

To evaluate the limit \(\lim_{n \to \infty} \left| \frac{n^{3/2}}{(n+1)^{3/2}} \right|\), we can divide both the numerator and denominator by \(n^{3/2}\), which simplifies it to \(\lim_{n \to \infty} \left| \frac{1}{(1+1/n)^{3/2}} \right|\). As n approaches infinity, \(1/n\) approaches 0, so the limit can be simplified further to \(1\).

Compare the limit with 1

The limit determined in the previous step is exactly 1. Therefore, according to the Ratio Test, the test is inconclusive about whether the series converges or diverges.