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Chapter 9: Problem 82

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{2}(n+2)}}$$

Short Answer

Expert verified
The series converges by the Limit Comparison Test with \( \sum \frac{1}{n^{3/2}} \).

Step by step solution

01

Understanding the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{2}(n+2)}} \). To determine convergence, we will apply convergence tests suitable for series with positive terms.
02

Simplify the Terms

Observe that \( \sqrt{n^2(n+2)} = \sqrt{n^3 + 2n^2} \). It is often useful to approximate complex terms. For large \( n \), \( n+2 \approx n \), so \( \frac{1}{\sqrt{n^3 + 2n^2}} \approx \frac{1}{n^{3/2}} \). This resembles a p-series.
03

Apply Comparison Test

A p-series of the form \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). For our series, the corresponding p-series is \( \sum \frac{1}{n^{3/2}} \), with \( p = 3/2 > 1 \). Therefore, the series \( \frac{1}{n^{3/2}} \) converges.
04

Conclude using Limit Comparison Test

To apply the Limit Comparison Test, use the series \( b_n = \frac{1}{n^{3/2}} \). Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n = \frac{1}{\sqrt{n^2(n+2)}} \). This becomes \( \lim_{n \to \infty} \frac{1}{\sqrt{n^3 + 2n^2}} \cdot n^{3/2} = 1 \). Since this limit is positive and finite, the convergence of \( \sum \frac{1}{n^{3/2}} \) implies that \( \sum \frac{1}{\sqrt{n^2(n+2)}} \) also converges.

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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 valuable tool in determining the convergence or divergence of series with non-negative terms. The idea is to compare the series in question to another series whose convergence is already known.

Here are the basic steps:
  • If the series \( \sum a_n \) can be compared with the series \( \sum b_n \), such that \( 0 \leq a_n \leq b_n \) for all points in the domain, and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
  • If \( \sum a_n \geq \sum b_n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.
To effectively use this test, you often need to find a comparable series that simplifies the term analysis, usually involving well-known series like the p-series or geometric series.
Limit Comparison Test
The Limit Comparison Test is similar to the Comparison Test but with a more sophisticated approach to establishing convergence. It involves calculating the limit of the ratio of two series' terms.

Here's how it works:
  • Choose a positive series \( \sum b_n \) that behaves similarly to \( \sum a_n \)
  • Compute the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \)
If \( L \) exists and is positive and finite, both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together. The key advantage of the Limit Comparison Test is that it can be easier to apply than directly bounding terms, as seen with the Comparison Test.
p-series
A p-series is a common mathematical series used as a benchmark for other series. It takes the form \( \sum \frac{1}{n^p} \). The convergence of a p-series is determined by the exponent \( p \).

  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
Understanding whether a complex series approximates or can be compared to a p-series is often the key to determining its convergence or divergence. This comparison is particularly useful in conjunction with series tests like the Comparison or Limit Comparison Tests.
Mathematical Series
A mathematical series is the sum of the terms of a sequence. Series are central to calculus and other branches of mathematics for understanding infinite processes.

For instance:
  • An arithmetic series involves repeated addition of a constant value.
  • A geometric series multiplies by a constant factor.
Convergence of a series refers to whether the infinite sum approaches a specific limit. This concept is critical for many applications in mathematics, enabling the precise modeling of various phenomena. Understanding series and their convergence is fundamental to analyzing complex sequences and functions.

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