/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 More sequences Find the limit of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

More sequences Find the limit of the following sequences or determine that the sequence diverges. $$\left\\{\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}\right\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence is 0 as n approaches infinity.

Step by step solution

01

Find the limit of the first term

Take the limit of $$\frac{75^{n-1}}{99^{n}}$$ as n goes to infinity. To simplify the expression, divide both the numerator and the denominator by 75^{n-1}: $$\lim_{n\to\infty} \frac{75^{n-1}}{99^{n}} = \lim_{n\to\infty} \frac{1}{\left(\frac{99}{75}\right)^{n}}$$ Now we can observe that the limit converges to 0, since the base of the power in the denominator is greater than 1. So, $$\lim_{n\to\infty} \frac{75^{n-1}}{99^{n}} = 0$$
02

Find the limit of the second term

Take the limit of $$\frac{5^{n} \sin n}{8^{n}}$$ as n goes to infinity. Since sine values (sin n) lie between -1 and 1, we can take their absolute values and find the limit: $$\lim_{n\to\infty} \frac{5^{n} |\sin n|}{8^{n}}$$ Divide both the numerator and denominator by 5^{n} to simplify the expression: $$\lim_{n\to\infty} \frac{|\sin n|}{\left(\frac{8}{5}\right)^{n}}$$ Again, the limit converges to 0, as the base of the power in the denominator is greater than 1. So, $$\lim_{n\to\infty} \frac{5^{n} \sin n}{8^{n}} = 0$$
03

Combine the results

The limits of both terms are 0, so we can combine these results to find the limit of the entire sequence: $$\lim_{n\to\infty} \left\\{\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}\right\\} = 0 + 0 = 0$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sequence Convergence
When we speak about sequence convergence, we refer to the behavior of a sequence as its terms progress toward infinity. If the terms of a sequence get closer and closer to a specific fixed value, we say that the sequence converges to that value or has a limit. For sequences, this means that for any given small positive number (no matter how small), there is a point beyond which all the terms of the sequence are within that distance from the limit. This concept is crucial for determining the stability and predictability of sequences in various fields like mathematics and science.In the original exercise, we examined terms like \(\frac{75^{n-1}}{99^{n}}\) and determined if they approach a single value as \(n\) grows. Through simplification and division by the highest power in the series, it was shown that the sequence indeed converges to a limit of 0. Recognizing convergence helps us make sense of complex sequences and understand their behavior at infinity.
Divergence of Sequences
Not all sequences converge to a single value. In some cases, the terms of a sequence do not approach any particular number, and such sequences are said to diverge. Divergence can happen in various forms, such as when a sequence infinitely increases or decreases, oscillates, or behaves chaotically.In our work with sequences like \(\frac{5^{n} \sin n}{8^{n}}\), despite the oscillating nature of \(\sin n\) (which swings between -1 and 1), the growth of the denominator \(8^n\) overpowers the sequence, pulling it consistently towards zero rather than a chaotic behavior. Therefore, understanding divergence involves checking if there's a predictable or stable pattern, or if the sequence fails to settle on a particular value. Getting to grips with these characteristics aids us in making decisions, particularly when working with infinite series and functions.
Infinite Limits
Infinite limits describe situations where the terms of a sequence grow indefinitely large as \(n\) increases. When we talk about the limit of a sequence being infinite, we're saying that the sequence doesn't approach a finite number but instead stretches towards positive or negative infinity.From our original problem, instead of observing an infinite limit, we saw that the terms diminished to zero, emphasizing convergence over an infinite outcome. However, understanding infinite limits remains essential, especially in higher mathematics, where they frequently model unbounded behaviors. Recognizing whether a sequence nears infinity or zero, as in our example, enables more accurate predictions and analyses of real-world phenomena involving sequences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether the following series converge. Justify your answers. $$\sum_{k=1}^{\infty} e^{-k^{3}}$$

Determine whether the following series converge. Justify your answers. $$\sum_{k=1}^{\infty}\left(\frac{1}{k^{2}}+\frac{1}{k^{5}}\right)$$

Prove that if \(\left\\{a_{n}\right\\} < \left\\{b_{n}\right\\}\) (as used in Theorem 10.6 ), then \(\left\\{c a_{n}\right\\} < \left\\{d b_{n}\right\\},\) where \(c\) and \(d\) are positive real numbers.

Determine whether the following series converge. Justify your answers. $$\sum_{k=1}^{\infty} \frac{\ln ^{2} k}{k^{3 / 2}}$$

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\) 4,...., 10. You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\}\), which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k .\) Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\) c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.