Test for convergence: $$\sum_{n=2}^{\infty} \frac{2 n^{3}}{n^{4}-2}$$

Short Answer

Expert verified
The series diverges by the Comparison Test.

Step by step solution

01

Identify the Series

The given series is \(\sum_{n=2}^{\infty} \frac{2 n^{3}}{n^{4}-2}\). We need to determine if this series converges.
02

Simplify the General Term

Simplify the general term \(a_n = \frac{2 n^{3}}{n^{4} - 2}\). For large values of \(n\), the term \(n^{4} - 2\) is dominated by \(n^{4}\), so the general term approximates to \(a_n \approx \frac{2 n^{3}}{n^{4}} = \frac{2}{n}\).
03

Use the Comparison Test

Use the Comparison Test to compare \(a_n\) with the harmonic series term \(\frac{2}{n}\). Note that \(\sum_{n=1}^{\infty} \frac{1}{n}\) (the harmonic series) diverges. We need to confirm if \(\sum_{n=2}^{\infty} \frac{2n^3}{n^4-2}\) has similar behavior.
04

Establishing the Comparison

Since \(\frac{2 n^{3}}{n^{4} - 2}\) is greater than \(\frac{2}{n+1}\) for \(n \ge 2\), and \(\sum_{n=2}^{\infty} \frac{2}{n}\) diverges, by the Comparison Test, the given series \(\sum_{n=2}^{\infty} \frac{2 n^{3}}{n^{4} - 2}\) also diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

comparison test
The Comparison Test is a handy tool to determine the convergence or divergence of a series by comparing it to a second series whose behavior is already known. To use it effectively, you follow these steps:
  • Identify the given series and simplify the general term if possible.
  • Find a second series whose convergence or divergence is known.
  • Compare the terms of the two series to decide if the original series converges or diverges.
The essential principle is that if your series has terms that are consistently larger than or comparable to a divergent series, your series will diverge as well. Conversely, if your series has terms that are consistently smaller than or comparable to a convergent series, your series will converge. In the given example, we compared \(\frac{2n^3}{n^4-2}\) with \(\frac{2}{n}\). Because we know that \(\frac{2}{n}\) (the harmonic series) diverges, and \(\frac{2n^3}{n^4-2}\) is larger than \(\frac{2}{n+1}\) for all .
harmonic series
The harmonic series is one of the fundamental examples in the study of infinite series. It is given by \(\textstyle\sum_{n=1}^{\infty} \frac{1}{n}\) and is known for its divergence. Despite its terms becoming smaller and smaller, the sum of the series grows without bound. Here are some key points about the harmonic series:
  • It diverges because it can be compared to the integral \(\textstyle\begin{align*}\textstyle\int_1^ x \frac{1}{t} dt\end{align*}\)
  • The divergence can be understood through the Comparison Test by splitting the harmonic series into manageable partial sums.
Understanding the properties of the harmonic series is crucial because it sets a precedent for evaluating more complex series like the one in the given exercise.
approximation of terms
Approximating terms is a common technique used to simplify the analysis of complex series. The idea is to focus on the behavior of the series as \ grows very large. Here’s how to approach it:
  • First, identify the leading term in the numerator and the denominator of the general term of the series.
  • For large values of \, ignore the smaller terms compared to the dominant term.
  • Use this approximate term to compare with known series for convergence or divergence analysis.
In this exercise, the term \(\textstyle\frac{2n^3}{n^4-2}\) simplifies to \(\textstyle\frac{2}{n}\) for large values of \. This approximation made it easier to compare our series with the harmonic series and thereby conclude its divergence. Approximations thus provide a streamlined method to handle otherwise complicated expressions.

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Most popular questions from this chapter

Test the following series for convergence. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} n}{n+5}$$

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. $$\sum_{0}^{\infty}(\sin x)^{n}(-1)^{n} 2^{n}$$

There are 9 one-digit numbers ( 1 to 9 ), 90 two-digit numbers ( 10 to 99 ). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\) are all greater than \(\frac{1}{10}\); similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series \(\left[\frac{1}{10}+\frac{1}{10}+\cdots\left(9 \text { terms each }=\frac{1}{10}\right)\right]+\left[90 \text { terms each }=\frac{1}{100}\right]+\cdots\) $$ =\frac{9}{10}+\frac{90}{100}+\cdots=\frac{9}{10}+\frac{9}{10}+\cdots $$ The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory \(m\) series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.

Use the integral test to show that \(\sum_{n=0}^{\infty} e^{-n^{2}}\) converges. Hint: Although you cannot evaluate the integral, you can show that it is finite (which is all that is necessary) by comparing it with \(\int^{\infty} e^{-n} d n\)

Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16 ). $$\sum_{n=2}^{\infty} \frac{1}{n \ln n}$$

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