Question

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

Short answer

Answer: The sequence \(a_n = n^{10}\) has a larger growth rate and overtakes the sequence \(b_n = n^9 \ln^3{n}\) at \(n = 7\).

Step-by-step solution

Compare the growth rate

To compare the growth rates, we need to find the ratio of \(a_n\) to \(b_n\): $$ \frac{a_n}{b_n} = \frac{n^{10}}{n^9 \ln^3{n}} $$ Now we need to simplify the expression.

Simplify the ratio

Divide the powers of \(n\): $$ \frac{a_n}{b_n} = \frac{n^{10}}{n^9 \ln^3{n}} = \frac{n}{\ln^3{n}} $$ Now we need to determine at which value of \(n\) this ratio is greater than 1.

Finding the value of n where the ratio is greater than 1

To find the value of n where \(a_n\) overtakes \(b_n\), we find n for which the ratio is greater than 1, that is: $$ \frac{n}{\ln^3{n}} > 1 $$ Multiplying both sides by \(\ln^3{n}\), $$ n > \ln^3{n} $$ Since \(n \ge 7\), we will find the smallest value of \(n \ge 7\) for which this inequality holds.

Solving the inequality

We can use trial and error to find the smallest value of \(n \ge 7\) for which \(n > \ln^3{n}\). You can either do this by hand or use software that computes values of expressions at different values of \(n\). Trying different values of \(n\): \(n = 7\): \(7 > \ln^3{7}\) as \(7 > 4.7682\) Therefore, the sequence \(a_n = n^{10}\) has a larger growth rate and overtakes the sequence \(b_n = n^9 \ln^3{n}\) at \(n = 7\).

Key concepts

Growth Rate of Sequences

Understanding the growth rate of sequences is essential for analyzing the long-term behavior of series in mathematics. In our exercise, we are introduced to two sequences, where the term growth rate refers to how quickly the values of the sequence increase as the index, or n, grows larger.

For sequences like a_n = n¹⁰ and b_n = n⁹ ln³n, the growth rate can determine which sequence will eventually become larger, regardless of their initial values. To compare these rates, we typically look at the behavior of the sequences as n approaches infinity. The sequence with terms that increase at a faster pace as n gets very large is said to have a higher growth rate.

In our specific case, the sequence with n¹⁰ grows faster than n⁹ ln³n because the polynomial term n¹⁰ will eventually outweigh the combined effect of the polynomial and logarithmic factors in the latter sequence as n becomes sufficiently large. This concept highlights the dominance of exponential growth over polynomial and logarithmic growth as values increase, which is a fundamental principle in sequences.

Ratio Comparison

The concept of ratio comparison is a powerful tool for analyzing and comparing the growth rates of sequences, which is evidenced in our problem where we are asked to compare a_n = n¹⁰ and b_n = n⁹ ln³n. By examining the ratio of the two sequences, we are essentially looking at how one sequence grows in relation to the other.

Here's how we apply this ratio comparison in the context of our sequences:

• We form the ratio \(\frac{a_n}{b_n} = \frac{n^{10}}{n^9 \ln^3{n}}\), which simplifies the comparison to \(\frac{n}{\ln^3{n}}\).
• The goal is to find when this ratio exceeds 1, which indicates that a_n is larger than b_n.

By simplifying the comparison to a single ratio, we can clearly see when one sequence outpaces the other, avoiding the complexity of comparing each term individually. This step is crucial in determining the point at which the growth rate of one sequence overtakes another.

Solving Inequalities

The final piece of the puzzle in our exercise is solving inequalities. This process is critical in determining the exact point at which the sequence a_n = n¹⁰ overtakes b_n = n⁹ ln³n. In general, inequalities are mathematical expressions that show the relationship between two values, indicating that one is greater than, less than, or equal to the other.

In our case, we focus on the inequality \(n > \ln^3{n}\). To solve this, we can utilize various methods, such as graphical analysis, numerical approximation, or analytical techniques like calculus. For students or problem solvers, numerical methods or trial and error can be most accessible:

• We test different values of n to see when the inequality becomes true.
• At n = 7, the inequality holds as \(7 > \ln^3{7}\).

This shows that by solving the inequality, we are able to find the threshold where one sequence surpasses the other in value. Mastering inequalities is essential for students because they form the basis of many mathematical analyses, including limits, optimization, and even real-life decision-making.