Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$$

Short Answer

Expert verified
The series diverges by the ratio test.

Step by step solution

01

- Identify the general term of the series

The general term of the given series is \(a_n = \frac{3^{2n}}{2^{3n}}\).
02

- Apply the ratio test formula

The ratio test requires finding the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
03

- Compute \(a_{n+1}\)

Substitute \(n+1\) into the general term: \(a_{n+1} = \frac{3^{2(n+1)}}{2^{3(n+1)}} = \frac{3^{2n+2}}{2^{3n+3}}\).
04

- Form the ratio \( \frac{a_{n+1}}{a_n} \)

Compute the ratio: \(\frac{a_{n+1}}{a_n} = \frac{\frac{3^{2n+2}}{2^{3n+3}}}{\frac{3^{2n}}{2^{3n}}} = \frac{3^{2n+2} \cdot 2^{3n}}{2^{3n+3} \cdot 3^{2n}} = \frac{3^2}{2^3} = \frac{9}{8}\).
05

- Take the limit

Since the ratio \(\frac{a_{n+1}}{a_n} \) is a constant \(\frac{9}{8}\), \(L = \frac{9}{8}\).
06

- Determine convergence or divergence

Apply the ratio test criterion: if \(L > 1\) the series diverges. Here, \( \frac{9}{8} > 1 \), thus the series diverges.

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 Vaia!

Key Concepts

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

Understanding Convergence
When studying infinite series, one of the key concepts is 'convergence'. A series converges if the sum of its terms approaches a specific finite number as the number of terms goes to infinity. To explain it simply, imagine trying to add up an endless list of numbers. If those numbers get smaller quickly enough, their total won't keep growing indefinitely. Instead, it'll level off near a particular value. For example, the series \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...\) converges to 1. This means that as you keep adding more and more terms of this series, the total gets closer and closer to 1, but never surpasses it. The ratio test helps us determine if a series converges by examining the ratio of successive terms in the series.
Exploring Divergence
Next, we have 'divergence', which is the opposite of convergence. A series diverges if the sum of its terms does not approach a specific finite value as the number of terms goes to infinity. In simpler words, if you keep adding more terms to the series and the total keeps growing without bounds, or wiggles endlessly without settling to a number, it diverges. For example, the series \(\frac{3^2}{2^3} + \frac{3^{4}}{2^{6}} + \frac{3^{6}}{2^{9}} + ...\) diverges because, as we found using the ratio test, the ratio of the successive terms is \(\frac{9}{8}\), which is greater than 1. This result tells us that each new term added makes the total keep increasing without bound, hence the series does not settle to any finite value.
Infinite Series Simplified
An 'infinite series' is simply an expression that represents adding an infinite sequence of terms. Consider it as trying to add an endless list. Some famous examples include geometric series and harmonic series. These infinite series can either converge to a number or diverge, based on the behavior of their terms. The concept of infinite series is crucial in various branches of mathematics and science as it helps model phenomena or solve problems involving continuous processes. The ratio test is a handy tool for determining the behavior of these series. It works by evaluating the limit of the ratio of successive terms. If the ratio turns out to be less than 1, the series converges. If more than 1, the series diverges. If equal to 1, the test is inconclusive, and further tests are needed.

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

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=0}^{\infty} \frac{(2 n) !}{3^{n}(n !)^{2}}$$

(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

The series \(\sum_{n=1}^{\infty} 1 / n^{s}, s>1,\) is called the Riemann Zeta function, \(\zeta(s) .\) \((\text { a })\) you found \(\zeta(2)=\pi^{2} / 6 .\) When \(n\) is an even integer, these series can be summed exactly in terms of \(\pi .\) ) By computer or tables, find $$\text { (a) } \quad \zeta(4)=\sum_{n=1}^{\infty} \frac{1}{n^{4}}$$ $$\text { (b) } \quad \zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$ $$\text { (c) } \quad \zeta\left(\frac{3}{2}\right)=\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}}$$

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\sin [\ln (1+x)]$$

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. $$\sum_{0}^{\infty} \frac{3^{n}(n+1)}{(x+1)^{n}}$$

See all solutions

Recommended explanations on Combined Science 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.

Sign-up for free