Problem 8
Question
Determine whether the series converges or diverse. $$\sum\left(\frac{2}{5}\right)^{k}$$
Step-by-Step Solution
Verified Answer
The given series is a geometric series with a common ratio \(r = \frac{2}{5}\). Since the absolute value of the common ratio is \(|\frac{2}{5}| = \frac{2}{5} < 1\), the series converges.
1Step 1: The given series is in the form of a geometric series: $$\sum\left(\frac{2}{5}\right)^{k}$$. Identify the common ratio, which is the value that is being multiplied to every consecutive term in the series. The common ratio in this series is \(r = \frac{2}{5}\). #Step 2: Check the convergence criterion#
In order to determine if the series converges or diverges, check the absolute value of the common ratio:
$$|r| = \left|\frac{2}{5}\right|$$
Since \(|\frac{2}{5}| = \frac{2}{5} < 1\), the series converges.
To conclude, the given series $$\sum\left(\frac{2}{5}\right)^{k}$$ converges.
2Step 2: Identify the type of series
Determine the series type and write out the first few terms if helpful.
3Step 3: Apply the appropriate test
Use the relevant convergence test, showing the computation.
4Step 4: Determine convergence/divergence or compute the sum
Conclude based on the test result.
5Step 5: State the final result
Clearly state the conclusion.
6Step 6: Conclude with the answer
The given series is a geometric series with a common ratio \(r = \frac{2}{5}\). Since the absolute value of the common ratio is \(|\frac{2}{5}| = \frac{2}{5} < 1\), the series converges.
Key Concepts
Common RatioConvergence CriteriaAbsolute Value in SeriesInfinite Series
Common Ratio
In the context of geometric series, the term 'common ratio' refers to the consistent factor that each term is multiplied by to get the next term in the sequence. Picture a row of dominoes lined up; each time one falls, it triggers the next one. Similarly, in a geometric sequence, every term generates the next one by being multiplied by the common ratio.
For instance, in the series \(\sum(\frac{2}{5})^{k}\), the common ratio is \(\frac{2}{5}\). If \(k\) represents each step in the series' progression, the next term is always \(\frac{2}{5}\) times its predecessor. This uniformity simplifies the process of analyzing series for properties like convergence.
For instance, in the series \(\sum(\frac{2}{5})^{k}\), the common ratio is \(\frac{2}{5}\). If \(k\) represents each step in the series' progression, the next term is always \(\frac{2}{5}\) times its predecessor. This uniformity simplifies the process of analyzing series for properties like convergence.
Convergence Criteria
When dealing with an infinite geometric series, the 'convergence criteria' is a set of conditions that determine whether the series adds up to a finite number or not. For a series to converge, the absolute value of its common ratio must be less than one. Why is this important? It's like filling a bucket with water using a cup that gets smaller with each pour. If each cup of water is less than the previous one, eventually you’ll be adding drops, and the bucket won’t overflow. Mathematically, this criterion ensures that the series' terms decrease quickly enough so that their sum approaches a specific limit.
If we apply this to our example, \(\left|\frac{2}{5}\right| = \frac{2}{5} < 1\), thus satisfying the convergence criteria, and leading us to conclude the series converges.
If we apply this to our example, \(\left|\frac{2}{5}\right| = \frac{2}{5} < 1\), thus satisfying the convergence criteria, and leading us to conclude the series converges.
Absolute Value in Series
The concept of 'absolute value' in series plays a key role in determining the behavior of the series. Absolute value, represented by two vertical lines \(|a|\), measures the distance of a number \(a\) from zero on the number line, without considering direction. This notion is crucial when we examine the convergence of series because it allows us to ignore the sign and focus on the magnitude of the terms.
In our geometric series example, the absolute value of the common ratio is crucial. The fact that \(\left|\frac{2}{5}\right| = \frac{2}{5}\) confirms the terms are getting smaller since \(\frac{2}{5}\) is less than 1. This decreasing magnitude is what ultimately causes the series to add up to a finite number, indicating convergence.
In our geometric series example, the absolute value of the common ratio is crucial. The fact that \(\left|\frac{2}{5}\right| = \frac{2}{5}\) confirms the terms are getting smaller since \(\frac{2}{5}\) is less than 1. This decreasing magnitude is what ultimately causes the series to add up to a finite number, indicating convergence.
Infinite Series
An 'infinite series' simply means that the number of terms in the series is endless—there's no final term. Imagine a road extending into the horizon without end; this is akin to an infinite series in mathematics. However, even with an infinite number of terms, some series can still have a finite total value. This seems counterintuitive at first – how can adding up an endless list of numbers give anything other than an infinitely large (or small) number?
This is where convergence comes into play. If the terms of an infinite series decrease rapidly enough, in accordance with the convergence criteria, the incremental additions eventually have a negligibly small impact, allowing a finite sum to emerge. The geometric series \(\sum(\frac{2}{5})^{k}\) exemplifies this by meeting the necessary conditions to converge, as shown in the previous sections.
This is where convergence comes into play. If the terms of an infinite series decrease rapidly enough, in accordance with the convergence criteria, the incremental additions eventually have a negligibly small impact, allowing a finite sum to emerge. The geometric series \(\sum(\frac{2}{5})^{k}\) exemplifies this by meeting the necessary conditions to converge, as shown in the previous sections.
Other exercises in this chapter
Problem 8
Test these series for (a) absolute convergence, (b) conditional convergence. \(\sum \frac{k^{3}}{2^{k}}\).
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Find the sum of the series. $$\sum_{k=0}^{\infty} \frac{1}{2^{k+3}}$$
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Evaluate. $$\sum_{k=1}^{5} \frac{(-1)^{2}}{k !}$$
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Determine whether the series converges or diverges. $$\sum \frac{1}{(\ln k)^{k}}$$
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