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Chapter 9: Problem 33

Use any method to determine whether the series converges. $$ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k(k+1)}} $$

Short Answer

Expert verified
The series converges by the Limit Comparison Test.

Step by step solution

01

Understand the Series Structure

Observe that the series is \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k(k+1)}} \). It is an infinite series with terms based on \( k \). Each term is a rational expression which involves a square root.
02

Analysis of Term Simplification

Let's examine the term \( \frac{1}{\sqrt{k(k+1)}} \). We notice that this can be expressed in terms of simpler expressions to facilitate testing for convergence.
03

Application of the Limit Comparison Test

We will compare this series with a simpler series using the Limit Comparison Test. Consider the series \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \). Note that \( \sqrt{k(k+1)} \approx \sqrt{k^2} = k \) as \( k \) becomes very large, so \( \frac{1}{\sqrt{k(k+1)}} \approx \frac{1}{k^{3/2}} \).
04

Compute the Limit for Comparison

Calculate the limit: \( \lim_{k \to \infty} \frac{\frac{1}{\sqrt{k(k+1)}}}{\frac{1}{k^{3/2}}} = \lim_{k \to \infty} \frac{k^{3/2}}{\sqrt{k(k+1)}} \). Simplify to get \( \lim_{k \to \infty} \frac{k^{3/2}}{k \sqrt{1+\frac{1}{k}}} = \lim_{k \to \infty} \frac{k^{3/2}}{k} \cdot \frac{1}{\sqrt{1 + \frac{1}{k}}} = \lim_{k \to \infty} \frac{k^{1/2}}{\sqrt{1 + \frac{1}{k}}} = 1 \). Since the limit equals 1, we can use the Limit Comparison Test.
05

Conclusion on Convergence

The series \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \) is a p-series with \( p = 3/2 > 1 \) and is known to converge. Given our limit comparison result, it indicates that \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k(k+1)}} \) also converges by the Limit Comparison Test.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a great tool in determining the convergence of an infinite series. Suppose you have a complicated series and want to know if it converges. Rather than working directly with it, you can compare it with another well-known, simpler series.
  • Choose a comparison series, often a p-series, where you can easily determine convergence or divergence.
  • Find the limit: \( \lim_{k \to \infty} \frac{a_k}{b_k} \), where \(a_k\) is the term in the original series and \(b_k\) is the term in the comparison series.
  • If this limit is a positive finite number, both series will converge or diverge together.
In our given problem, the series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k(k+1)}} \) was compared to \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \). By simplifying and calculating the limit, it was found that their ratio converges to 1, indicating they behave similarly. Thus, the original series converges as well.
P-series
A p-series is a type of series formed as \( \sum_{k=1}^{\infty} \frac{1}{k^p} \). In this series, the convergence depends on the value of the exponent \( p \).
  • If \( p > 1 \), the p-series converges.
  • If \( p \leq 1 \), the p-series diverges.
In our exercise, the comparison series \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \) is a p-series where \( p = 3/2 \). Since 3/2 is greater than 1, this series is known to converge. This fundamental property helps in determining the behavior of more complex series by employing it in a limit comparison, as done in this problem.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. Convergence of such a series means you can keep adding terms indefinitely, but the sum approaches some finite number.
  • To determine if a series converges, you have various tests available like comparison tests, ratio tests, and the Limit Comparison Test.
  • Understanding and simplifying each term is often crucial in these checks.
  • Many infinite series have terms that behave similarly to well-known series, aiding in easier convergence testing.
The series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k(k+1)}} \) is infinite with terms that asymptotically resemble those of the p-series \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \). Through the Limit Comparison Test, this relationship helped confirm its convergence, demonstrating the principle of comparing complex infinite series to simple benchmarks.

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Most popular questions from this chapter

Suppose that the power series \(\sum c_{k}\left(x-x_{0}\right)^{k}\) has radius of convergence \(R\) and \(p\) is a nonzero constant. What can you say about the radius of convergence of the power series \(\sum p c_{k}\left(x-x_{0}\right)^{k}\) ? Explain your reasoning. [Hint: See Theorem \(9.4 .3 .]\)

Find the radius of convergence and the interval of convergence. $$ \sum_{k=0}^{\infty} \frac{k !}{2^{k}} x^{k} $$

(a) Find examples to show that if \(\sum a_{k}\) converges, then \(\sum a_{k}^{2}\) may diverge or converge. (b) Find examples to show that if \(\sum a_{k}^{2}\) converges, then \(\quad \sum a_{k}\) may diverge or converge.

Apply the divergence test and state what it tells you about the series. $$ \begin{array}{ll}{\text { (a) } \sum_{k=1}^{\infty} \frac{k^{2}+k+3}{2 k^{2}+1}} & {\text { (b) } \sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{k}} \\\ {\text { (c) } \sum_{k=1}^{\infty} \cos k \pi} & {\text { (d) } \sum_{k=1}^{\infty} \frac{1}{k !}}\end{array} $$

Determine whether the series converges. $$ \sum_{k=1}^{\infty} \operatorname{sech}^{2} k $$
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