Chapter 1: Q8MP (page 1)

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

Short Answer

Expert verified

The limit is greater than 0, so the series $\sum_{n = 2}^{\infty} \frac{2 n^{3}}{n^{4} - 2}$ diverges.

Step by step solution

Concept and formula used to show that the series diverges

A special comparison test has two parts as follows:

1- The series $\sum_{n = 1}^{\infty} a_{n}$ converges if $a_{n} ⩾ 0$ and $\frac{a_{n}}{b_{n}}$ tends to a finite limit.

2- The series $\sum_{n = 1}^{\infty} a_{n}$ diverges if $a_{n} ⩾ 0$ and $\frac{a_{n}}{b_{n}}$ tends to a limit greater than 0.

Calculation to show that the given series diverges

Let $a_{n} = \frac{2 n^{3}}{n^{4} - 2}$ .

Consider the given series as comparison series as follows:

$\sum_{n = 2}^{\infty} \frac{n^{3}}{n^{4}} = \sum_{n = 2}^{\infty} \frac{1}{n}$

Take

$b_{n} = \frac{1}{n}$

Here, it is known that $\sum b_{n} = \sum \frac{1}{n}$ is a divergent series.

Now, use the special comparison test and simplify as follows:

role="math" localid="1664274810995" $l i m_{n \to \infty} \frac{a_{n}}{b_{n}} = l i m_{n \to \infty} \frac{a_{n}}{b_{n}} (\frac{\frac{2 n^{3}}{n^{4} - 2}}{\frac{1}{n}}) = l i m_{n \to \infty} \frac{a_{n}}{b_{n}} (\frac{2 n^{3}}{n^{4} (1 - \frac{2}{n^{4}})} \cdot n) = \frac{2}{1 - \frac{2}{n^{4}}} = 2$

Here, the $l i m_{n \to \infty} \frac{a_{n}}{b_{n}} = 2 > 0$ . So, the given series diverges.

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

Short Answer

Step by step solution

Concept and formula used to show that the series diverges

Calculation to show that the given series diverges

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Test for convergence: ∑n=2∞2n3n4-2

Short Answer

Step by step solution

Concept and formula used to show that the series diverges

Calculation to show that the given series diverges

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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