Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{\ln (1 / n)}{\sqrt{2 n}} $$

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Plug values into formula

First, substitute the values of \( n \) from 1 to 5 into the formula \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \) to find the first five terms of the sequence.

Step 2: Calculate the first term

Substitute \( n = 1 \): \(a_1 = \frac{\ln(1/1)}{\sqrt{2 \times 1}} = \frac{0}{\sqrt{2}} = 0 \) So, the first term is 0.

Step 3: Calculate the second term

Substitute \( n = 2 \): \(a_2 = \frac{\ln(1/2)}{\sqrt{2 \times 2}} = \frac{\ln(0.5)}{2} \approx \frac{-0.693}{2} = -0.3465 \)So, the second term is approximately -0.3465.

Step 4: Calculate the third term

Substitute \( n = 3 \): \(a_3 = \frac{\ln(1/3)}{\sqrt{2 \times 3}} = \frac{\ln(0.333)}{\sqrt{6}} \approx \frac{-1.099}{2.449} = -0.4485 \)So, the third term is approximately -0.4485.

Step 5: Calculate the fourth term

Substitute \( n = 4 \): \(a_4 = \frac{\ln(1/4)}{\sqrt{2 \times 4}} = \frac{\ln(0.25)}{2 \sqrt{2}} \approx \frac{-1.386}{2.828} = -0.4906\)So, the fourth term is approximately -0.4906.

Step 6: Calculate the fifth term

Substitute \( n = 5 \): \(a_5 = \frac{\ln(1/5)}{\sqrt{2 \times 5}} = \frac{\ln(0.2)}{3.162} \approx \frac{-1.609}{3.162} = -0.5088 \)So, the fifth term is approximately -0.5088.

Step 7: Analyze convergence

To determine whether the sequence converges, we need to analyze the behavior of \( a_n \) as \( n \to \infty \). Notice that \( \ln(1/n) = -\ln(n) \), so as \( n \) increases, \( \ln(n) \to \infty \), making \( \ln(1/n) \to -\infty \). However, since the \( \sqrt{2n} \) grows faster than \( \ln(n) \), the sequence's terms will grow nearer to zero, suggesting it converges to zero.

Step 8: Conclusion

The sequence \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \) converges, and the limit as \( n \rightarrow \infty \) is 0.

Key Concepts

Limits

When studying sequences, the concept of limits is essential in determining if a sequence converges to a particular value. In this exercise, we examined the sequence given by the formula \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \). To find if the sequence converges, we analyze what happens to \( a_n \) as \( n \rightarrow \infty \).

The limit of a sequence as \( n \) approaches infinity reveals the value that the sequence approaches. For this sequence, we need to determine if \( \lim_{n \to \infty} a_n = 0 \).

• The numerator, \( \ln(1/n) \), simplifies to \(-\ln(n)\) because \( \ln(1/n) = -\ln(n) \).
• As \( n \) increases, \( \ln(n) \) grows larger, pushing \(-\ln(n)\) towards \(-\infty\).
• The denominator, \( \sqrt{2n} \), grows even faster because the square root of a number that is multiplying by 2 becomes greater with each incremental increase in \( n \).

Combining these observations, the fraction becomes smaller and smaller, approaching zero. Hence, we conclude that the limit of the sequence is 0.

Logarithmic Functions

Logarithmic functions are critical in analyzing sequences and functions involving exponential growth or decay. In this problem, the sequence \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \) prominently features the logarithm as part of the expression. The natural logarithm function, \( \ln(x) \), is the inverse of the exponential function \( e^x \).

Understanding the properties of logarithms is crucial:

• \( \ln(1/n) = -\ln(n) \) shows that the logarithm of a reciprocal becomes the negative logarithm of the number itself.
• The natural logarithm, \( \ln(n) \), grows slowly as \( n \) increases, compared to other functions like polynomials or exponentials.
• Since \( \ln(n) \rightarrow \pm \infty \) much slower than polynomials, it is often part of convergence analysis.

In this sequence, recognizing that \( \ln(1/n) = -\ln(n) \) helps simplify the approach to determining the behavior of \( a_n \) as \( n \) grows large.

Square Roots

Square roots are often used in mathematical expressions to manage growth rates or simplify equations. In the sequence \( a_n = \frac{\ln(1/n)}{\sqrt{2n}} \), the square root plays a critical role in moderating the growth of the denominator as \( n \) increases.

The square root function, \( \sqrt{x} \), effectively slows down the growth of \( x \):

• While \( n \) doubles, \( \sqrt{n} \) only increases by a factor of approximately \( \sqrt{2} \).
• This slower growth is useful in convergence analysis, as it can counterbalance faster-growing numerators, leading to a convergent sequence.
• In the exam sequence, \( \sqrt{2n} \) is the dominant term, ensuring that even though the numerator grows negatively, the entire fraction trends towards zero.

By understanding this balancing act, we see why sequences involving both logarithms and square roots often converge, especially when the square root term grows faster than the logarithmic term in the numerator.