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Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}$$

Short Answer

Expert verified
$$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{p^k k!}$$ Answer: The series converges for all values of p > 2.

Step by step solution

01

Rewrite the series using the given formula

We are given the following series to investigate convergence: $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{p^k k!}$$
02

Apply the Ratio Test

We will now apply the Ratio Test to find the limit of the ratio of consecutive terms in the series. Recall that for the Ratio Test, we take the limit as k approaches infinity of the ratio of consecutive terms: $$\lim_{k \to \infty} \frac{\frac{1 \cdot 3 \cdot 5 \cdots (2(k+1)-1)}{p^{k+1} (k+1)!}}{\frac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{p^k k!}}$$
03

Simplify the limit expression

To simplify this limit expression, note that: $$1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2(k+1)-1) = 1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1)$$ Now, we can simplify the limit as: $$\lim_{k \to \infty} \frac{\frac{1 \cdot 3 \cdot 5 \cdots (2k+1)}{p^{k+1} (k+1)!}}{\frac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{p^k k!}} = \lim_{k \to \infty} \frac{(2k+1)p^k k!}{p^{k+1}(k+1)!}$$
04

Simplify further and compute the limit

The expression inside the limit can be simplified further by canceling out terms: $$\lim_{k \to \infty} \frac{(2k+1)p^k k!}{p^{k+1}(k+1)!} = \lim_{k \to \infty} \frac{(2k+1)}{p(k+1)}$$ Now, we can compute the limit easily: $$\lim_{k \to \infty} \frac{(2k+1)}{p(k+1)} = \frac{2}{p}$$
05

Determine the values of p for which the series converges

According to the Ratio Test, if the limit is less than 1, the series converges, and if the limit is greater than 1, the series diverges. Thus, we must solve the inequality: $$\frac{2}{p} < 1$$ To solve this inequality, multiply both sides by p (since p > 0, it doesn't change the inequality sign): $$2 < p$$ This result tells us that the series converges for all values of p > 2.

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Key Concepts

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

Ratio Test
The Ratio Test is a powerful technique used to determine the convergence of an infinite series. It's especially helpful for series with complex terms, such as factorials or powers. To apply the Ratio Test, we examine the ratio of consecutive terms in a series.
The formula for the Ratio Test involves taking the limit:
  • Calculate the ratio of the \(k+1\)th term to the \(k\)th term.
  • Take the limit of this ratio as \(k\) approaches infinity.
  • If the limit is less than 1, the series converges absolutely.
  • If the limit is greater than 1, the series diverges.
  • If the limit equals 1, the test is inconclusive.
In practice, this method simplifies complex series to a manageable limit expression, allowing us to determine convergence or divergence based on a simple comparison.
Series expansion
Series expansion involves expressing a sequence or function as a sum of terms. This is essential in tackling complex mathematical problems because it breaks down functions or other entities into simpler, more manageable parts. In the given exercise, a series involving an odd product \(1 \cdot 3 \cdot 5 \cdots (2k-1)\) is expanded per term to analyze convergence.
To simplify such series, we need to:
  • Recognize the pattern or formula generating the terms.
  • Write the formula for the general term.
  • Simplify or manipulate this general term to facilitate convergence testing, often by using the Ratio Test or similar methods.
Series expansion is foundational for solving many convergence problems, as it allows the identification of patterns and behaviors in infinite sequences.
Factorial notation
Factorial notation, denoted by an exclamation mark (!), is a mathematical shorthand used to represent the product of all positive integers up to a given number. For instance, \(n!\) means \(n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\). In the series provided, factorials appear in the denominator, representing sequences that grow very fast.
Here's how factorial notation is applied in our exercise:
  • The factorial \(k!\) means multiplying all integers from 1 to k, making it a rapid-growing sequence.
  • When analyzing series, factorial terms often need to be managed by simplification techniques such as the Ratio Test.
  • Understanding how to handle factorials is crucial because they can significantly impact the convergence of a series due to their size and growth rate.
In many convergence problems, factorials appear in denominators, thus influencing the series towards convergence rather than divergence.
Inequality solving
Inequality solving is a method for determining ranges or conditions under which a mathematical statement holds true. In the context of series convergence, we used inequations to figure out valid values for parameters.
Specifically:
  • Once the Ratio Test is applied, we obtain a limit inequality \((\frac{2}{p} < 1)\).
  • Solving this inequality involves basic algebraic steps, like isolating the variable (in this case, p) by rearranging and simplifying terms.
  • The solution tells us which values of p allow the series to converge (here, \(p > 2\)).
Inequality solving thereby offers a direct method to fine-tune and restrict parameters within a mathematical model, crucial for determining convergence conditions.

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Most popular questions from this chapter

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Determine whether the following statements are true and give an explanation or counterexample. a. A series that converges must converge absolutely. b. A series that converges absolutely must converge. c. A series that converges conditionally must converge. d. If \(\sum a_{k}\) diverges, then \(\Sigma\left|a_{k}\right|\) diverges. e. If \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) converges. f. If \(a_{k}>0\) and \(\sum a_{k}\) converges, then \(\Sigma a_{k}^{2}\) converges. g. If \(\Sigma a_{k}\) converges conditionally, then \(\Sigma\left|a_{k}\right|\) diverges.

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