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Magazine Chapter 9: Problem 10
Use the limit comparison test to determine whether the series converges. \(\sum_{k=1}^{\infty} \frac{1}{(2 k+3)^{17}}\)
Short Answer
Expert verified
The series \( \sum_{k=1}^{\infty} \frac{1}{(2 k+3)^{17}} \) converges.
Step by step solution
01
Identify the Series
The given series is \( \sum_{k=1}^{\infty} \frac{1}{(2 k+3)^{17}} \). Our task is to determine if this series converges using the limit comparison test.
02
Choose a Comparison Series
We will compare the given series to \( \sum_{k=1}^{\infty} \frac{1}{k^{17}} \), a p-series with \( p = 17 \). We know that a p-series converges if \( p > 1 \).
03
Apply the Limit Comparison Test
The limit comparison test requires us to compute \( \lim_{k \to \infty} \frac{a_k}{b_k} \), where \( a_k = \frac{1}{(2k+3)^{17}} \) and \( b_k = \frac{1}{k^{17}} \).
04
Compute the Limit
Calculate \( \lim_{k \to \infty} \frac{\frac{1}{(2k+3)^{17}}}{\frac{1}{k^{17}}} = \lim_{k \to \infty} \frac{k^{17}}{(2k+3)^{17}} \). Simplify this expression: \[ \lim_{k \to \infty} \left( \frac{k}{2k+3} \right)^{17} = \left( \lim_{k \to \infty} \frac{k}{2k+3} \right)^{17} = \left( \frac{1}{2} \right)^{17} = \frac{1}{2^{17}}. \]
05
Analyze the Computed Limit
The limit calculated in the previous step is non-zero and finite: \( \frac{1}{2^{17}} > 0 \). According to the limit comparison test, since the original series \( \sum \frac{1}{k^{17}} \) converges, the series \( \sum \frac{1}{(2k+3)^{17}} \) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergent Series
A convergent series is essentially an infinite series whose terms approach a finite number. Think of a series as a sum of an infinite list of numbers. When we say it "converges," we mean that if you kept adding terms forever, you would get closer and closer to a specific number. This is very interesting because even though there are infinite terms, we can still end up with a finite sum. This is a key concept in calculus and is fundamental in understanding the behavior of series.
- Convergence occurs when the sum of an infinite series results in a finite value.
- In calculating convergence, each term individually gets closer to zero as you keep adding them to see if there is a limit.
- Strict mathematical tests, like the Limit Comparison Test, are used to confirm a series is convergent.
P-Series Test
The P-Series Test is a powerful tool in determining whether a series converges or diverges. A p-series is of the form \[ \sum_{k=1}^{\infty} \frac{1}{k^p} \]where \( p \) is a positive constant. The beauty of the P-Series Test lies in its simplicity: the series converges if \( p > 1 \), and diverges if \( p \leq 1 \).
Let's see why this is useful:
Let's see why this is useful:
- If you know the value of \( p \), you can immediately determine convergence or divergence without complicated calculations.
- It helps compare complex series to a simpler p-series to infer convergence properties.
- By understanding the P-Series, you can gain insights into many related series using comparison tests.
Infinite Series
Infinite series are sequences of numbers added together in an ongoing manner. They can be daunting because they never stop - there are always more terms to consider. However, they are critically important in mathematics because they can represent everything from tiny measurements in physics to entire functions in calculus.
With infinite series, the goal is often to find a beautiful balance in mathematics, where even an endless process can lead to a satisfying and finite solution.
- An infinite series is written as \( \sum_{k=1}^{\infty} a_k \), where \( a_k \) are the terms of the series.
- A key question is whether the series converges (adds up to a finite number) or diverges (grows indefinitely).
- Tests like the Limit Comparison Test help us understand whether an infinite series will converge or diverge.
With infinite series, the goal is often to find a beautiful balance in mathematics, where even an endless process can lead to a satisfying and finite solution.
