Question

Use the comparison test to determine whether the following series converge. $$ \sum_{n=1}^{\infty} \frac{\sqrt[4]{n}}{\sqrt[3]{n^{4}+n^{2}}} $$

Short answer

The series diverges by the Comparison Test.

Step-by-step solution

Simplify the Given Series

First, express the given series \[\sum_{n=1}^{\infty} \frac{\sqrt[4]{n}}{\sqrt[3]{n^{4}+n^{2}}}\]in a simpler form. The numerator can be rewritten as \(n^{1/4}\), and the denominator inside the cube root can be approximated by the dominant term, \(n^4\), for large \(n\):\[\sqrt[3]{n^4 + n^2} \approx (n^4)^{1/3} = n^{4/3}.\]Thus, the term \(a_n\) can be simplified to approximately:\[a_n \approx \frac{n^{1/4}}{n^{4/3}} = n^{1/4 - 4/3} = n^{-11/12}.\]

Choose a Series to Compare

Choose a comparison series that resembles \(n^{-11/12}\). A natural choice is the p-series:\[\sum_{n=1}^{\infty} \frac{1}{n^{11/12}}.\]The p-series \(\sum \frac{1}{n^p}\) converges if \(p > 1\) and diverges otherwise. Here, since \(11/12 < 1\), this series diverges.

Apply the Comparison Test

To apply the Comparison Test, verify that for sufficiently large \(n\), the terms of the original series \(a_n\) satisfy \[a_n \geq \frac{1}{n^{11/12}}.\]Simplifying the expression for \(a_n\) obtained in Step 1, and comparing it with the chosen p-series, we find that:\[\frac{\sqrt[4]{n}}{\sqrt[3]{n^4+n^2}} \geq \frac{n^{1/4}}{n^{4/3}} = \frac{1}{n^{11/12}}.\]Since both series terms maintain this inequality for large \(n\), and the comparison series diverges, the original series also diverges.

Key concepts

Comparison Test

The Comparison Test is a handy tool when faced with tricky series, like those involving roots or complex expressions. It leverages our understanding of simpler or more recognizable series to draw conclusions about more complicated ones.

• The basic idea is to compare your series with another series whose convergence behavior (converges or diverges) is already known.
• It's crucial for the inequality to hold for all sufficiently large terms of the series. For convergence, you would show that your series is less than or equal to a convergent series.
• For divergence, like in the exercise, you show your series is greater than or equal to a series that is known to diverge.

When using the comparison test, especially for a series like \( \sum_{n=1}^{\infty} \frac{\sqrt[4]{n}}{\sqrt[3]{n^{4}+n^{2}}} \), you simplify the terms first. This helps in identifying a comparable series.

Series Divergence

When a series does not meet the criteria for convergence, it diverges. Understanding series divergence is fundamental in analysis because it tells us that the sum does not approach a finite value.
For the series examined in the exercise, the terms simplify to mimic a p-series,\[ \sum_{n=1}^{\infty} \frac{1}{n^{11/12}}. \]A p-series diverges if its exponent \(p\) is less than or equal to 1. In this case,\(p = 11/12\), so the series diverges.It highlights the importance of recognizing series that resist collapsing into a finite number as you sum an increasing number of terms. It's a useful watchpoint when applying the comparison test. If the comparator diverges and is approached or exceeded by our target series, divergence is maintained.

Dominant Term Approximation

The concept of dominant term approximation simplifies complex expressions, especially when dealing with polynomials or rational expressions in series.
For a large \(n\), the higher-power terms significantly dominate the expression. In the original exercise, \[ \frac{\sqrt[4]{n}}{\sqrt[3]{n^{4}+n^{2}}} \] is approximated with \( n^{-11/12} \), by noticing that \(n^4\) is the dominant term in \(n^4 + n^2.\)This approximation is crucial because manipulating simpler terms allows us to use known tests and series. It reduces unnecessary complexities and focuses on terms that influence behavior as \(n\) grows large.When making an approximation, carefully analyze which term repels the rest, so contrast and comparison tests become feasible.