Show that \(\sum_{n=2}^{\infty} 1 / n^{3 / 2}\) is convergent. What is wrong with the following "proof" that it diverges? $$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots$$ which is $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\dots=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$. since the harmonic series diverges, the original series diverges. Hint: Compare \(3 n\) and \(n \sqrt{n}\).

The p-series test is a fundamental tool in determining the convergence or divergence of a series. A series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) is known as a p-series. This test states that the series converges if and only if the exponent \(p \) is greater than 1.

To understand why the p-series test is helpful, consider the behavior of the terms as \(n \) becomes very large. When \(p > 1 \), the terms \( \frac{1}{n^p} \) get very small very quickly, making the sum of the terms finite. Conversely, if \( p \leq 1 \), the terms decrease too slowly, leading to an infinite sum.

For the given problem, we analyzed the series \(\sum_{n=2}^{\infty} \frac{1}{n^{3/2}}\). Here, our \(p \) value is 1.5, which is greater than 1. This means our series converges according to the p-series test.

The harmonic series is one of the most famous divergent series. It is given by \(\sum_{n=1}^{\infty} \frac{1}{n}\). Despite the terms getting smaller as \(n \) increases, the series sums to infinity.

This is quite interesting because, at first glance, the terms \( \frac{1}{n} \) shrink, but not quickly enough to sum to a finite value. Instead, it grows without bound.

In the given incorrect proof in the exercise, the harmonic series was used as a comparison to argue that \(\sum_{n=2}^{\infty} \frac{1}{n^{3/2}}\) diverges. However, this was invalid as the convergence or divergence of a series must be tested with appropriate comparisons. The harmonic series diverges, but it doesn't serve as a suitable comparator for a p-series with \( p > 1\).

Making correct comparisons between series is crucial for determining their convergence or divergence. In the discussed exercise, an attempt was made to compare \(\sum_{n=2}^{\infty} \frac{1}{n^{3/2}}\) with terms related to the harmonic series.

This comparison was flawed because it didn't consider the correct nature of each series. When using comparisons, it is essential to match the series with appropriate counterparts – in this case, comparing a convergent p-series against the diverging harmonic series was incorrect.

The appropriate way to correctly compare series is by understanding the behavior of the terms and applying relative convergence tests. For instance, comparing \(\frac{1}{n^{3/2}}\) with a series where \(p > 1\), consistently will show convergence, establishing that the original series \(\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\) converges. Using improper comparisons results in invalid conclusions, emphasizing the need for precise mathematical reasoning.