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Chapter 9: Problem 18

In Problems \(1-20,\) an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\},\) determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{\ln (1 / n)}{\sqrt{2 n}} $$

Short Answer

Expert verified
The sequence converges to 0.

Step by step solution

01

Calculate the First Term

To find the first term of the sequence, substitute \(n = 1\) into the formula: \[ a_1 = \frac{\ln(1/1)}{\sqrt{2 \times 1}} = \frac{\ln(1)}{\sqrt{2}} = \frac{0}{\sqrt{2}} = 0 \]
02

Calculate the Second Term

Substitute \(n = 2\) into the formula to find the second term: \[ a_2 = \frac{\ln(1/2)}{\sqrt{2 \times 2}} = \frac{\ln(0.5)}{2} = \frac{-0.6931}{2} \approx -0.3466 \]
03

Calculate the Third Term

Substitute \(n = 3\) into the formula: \[ a_3 = \frac{\ln(1/3)}{\sqrt{2 \times 3}} = \frac{\ln(0.3333)}{\sqrt{6}} = \frac{-1.0986}{\sqrt{6}} \approx -0.4487 \]
04

Calculate the Fourth Term

Substitute \(n = 4\) into the formula: \[ a_4 = \frac{\ln(1/4)}{\sqrt{2 \times 4}} = \frac{\ln(0.25)}{2 \sqrt{2}} = \frac{-1.3863}{2 \sqrt{2}} \approx -0.4901 \]
05

Calculate the Fifth Term

Substitute \(n = 5\) into the formula: \[ a_5 = \frac{\ln(1/5)}{\sqrt{2 \times 5}} = \frac{\ln(0.2)}{\sqrt{10}} = \frac{-1.6094}{\sqrt{10}} \approx -0.5099 \]
06

Determine Convergence or Divergence

Examine the behavior of the terms as \(n\) increases. The sequence formula is:\[ a_n = \frac{\ln(1/n)}{\sqrt{2n}} = \frac{-\ln(n)}{\sqrt{2n}} \]As \(n\) approaches infinity, both \(\ln(n)\) and \(\sqrt{2n}\) grow larger, but the logarithmic growth is slower than the square root growth, leading \(a_n\) to approach 0.
07

Calculate the Limit

Use L'H么pital's Rule to evaluate the limit:The limit of \( \frac{-\ln(n)}{\sqrt{2n}} \) as \(n\) approaches infinity is indeterminate of the form \(\frac{-\infty}{\infty}\). Differentiate the numerator and denominator:\[ f(n)=\ln(n), \, f'(n)=\frac{1}{n}, \quad g(n)=\sqrt{2n}, \, g'(n)=\frac{1}{\sqrt{8n}} \] Now applying L'H么pital's Rule:\[ \lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2n}} = \lim_{n \to \infty} \frac{-\frac{1}{n}}{\frac{1}{2\sqrt{2n}}} = \lim_{n \to \infty} \frac{-2\sqrt{2n}}{n} = \lim_{n \to \infty} \frac{-2\sqrt{2}}{\sqrt{n}} = 0 \]Thus, the sequence converges and the limit is 0.

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Key Concepts

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

The Explicit Formula of a Sequence
When we talk about an explicit formula for a sequence, we refer to a rule that allows you to find the value of any term in a sequence directly from its position number. This is different from a recursive formula, which requires you to know previous terms. In the given exercise, the explicit formula for the sequence is \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \). What this means is:
  • You can plug in any value for \( n \) to calculate the \( n^{th} \) term of the sequence.
  • This formula is used to compute the values sequentially for \( n = 1, 2, 3, \) and so on.
To know if the formula makes the sequence converge or diverge, we also need additional analysis, which involves understanding how the terms behave as \( n \) gets very large.
Understanding L'H么pital's Rule
L'H么pital's Rule is a popular technique in calculus for finding limits that present indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It simplifies the process of finding these limits by allowing differentiation of the numerator and denominator until the limit is no longer indeterminate.
  • In our sequence, \( a_n = \frac{-\ln(n)}{\sqrt{2n}} \), as \( n \rightarrow \infty \), the form is \( \frac{-\infty}{\infty} \), which is indeterminate.
  • By applying L'H么pital's Rule, we differentiate the numerator \( (-\ln(n)) \) to get \( \frac{-1}{n} \), and the denominator \( (\sqrt{2n}) \) to get \( \frac{1}{2\sqrt{2n}} \).
  • After differentiation, the limit becomes solvable, as evidenced by the finite computations revealing it approaches zero.
Using L'H么pital's Rule effectively requires knowing how to differentiate these kinds of functions.
Sequence Limit Calculation
Finding the limit of a sequence as \( n \rightarrow \infty \) is often crucial for understanding its long-term behavior. In this particular problem, we needed to establish whether the sequence \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \) converges, and if so, what it converges to.
  • First, observe that the sequence converges if the limit exists and is finite.
  • Since \( \ln(n) \) grows slower than \( \sqrt{2n} \), one can anticipate that the fraction \( \frac{-\ln(n)}{\sqrt{2n}} \) diminishes near zero as \( n \) becomes very large.
  • L'H么pital's Rule provided the mathematical means to calculate that limit precisely, showing that the entire sequence tended toward zero.
Sequence limit calculation is a pivotal concept as it tells you the ultimate value the terms of the sequence are approaching.

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Most popular questions from this chapter

Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. \(1+2 x+2^{2} x^{2}+2^{3} x^{3}+2^{4} x^{4}+\cdots\)

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{1}{1-\sin x} $$

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Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=x \sec \left(x^{2}\right)+\sin x $$

Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12)\). $$ f(x)=\sqrt{1+x} $$
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