Question

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 \(n \sqrt{n}\).

Short answer

The series converges by the p-series test because \( p = \frac{3}{2} \), and the comparison to the harmonic series is incorrect because the terms \(\frac{1}{n^{3/2}}\) do not correspond to a diverging harmonic series.

Step-by-step solution

State the Series

The series in question is \[\sum_{n=2}^{\infty} \frac{1}{n^{3/2}}\]

Use the p-Series Test for Convergence

A series of the form \[\sum_{n=1}^{\infty} \frac{1}{n^{p}}\] converges if and only if \( p > 1 \). Here, the given series has \( p = \frac{3}{2} \), which is greater than 1. Therefore, the series converges.

Analyze the Incorrect Proof

The given proof compares terms from the original series with terms from another series and argues divergence based on the harmonic series. Specifically, it states: \[\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots > \frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots\]

Check Comparison Series

The proof incorrectly claims that: \[\frac{1}{3} + \frac{1}{6} + \frac{1}{9} + \frac{1}{12} + \cdots = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)\] is a comparison to the harmonic series, which diverges. The key point to note here is that the harmonic series does not compare correctly to the terms \(\frac{1}{n^{3/2}}\); rather, the inequality and comparison provided are incorrect.

Use Proper Series Comparison

To correctly determine the nature of our series, compare it to a known convergent series. Since \(\frac{1}{n^{3/2}}\) is a p-series with \(p > 1\), it inherently converges per the p-series test, making any comparison to a diverging series like the harmonic series inappropriate.

Key concepts

p-series test

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.

harmonic series

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 \( p > 1\).

improper comparisons in series

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.