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Determine whether the series is absolutely convergent, conditionally convergent or divergent. $$\sum_{k=0}^{\infty}(-1)^{k} \frac{3}{k !}$$

Short Answer

Expert verified
The given series is absolutely convergent, which implies it's also conditionally convergent.

Step by step solution

01

Test for Absolute Convergence

Apply the Ratio Test, which states that if the limit as n approaches infinity of the absolute value of the (n+1)th term divided by the nth term is less than 1, the series converges absolutely. For our series, the (n+1)th term is \(3/(k+1)!\) and the nth term is \(3/k!\). Take the absolute value and find the limit as k approaches infinity: \[\lim_{k \rightarrow \infty} \left| \frac{3/(k+1)!}{3/k!} \right|\] This simplifies to: \[\lim_{k \rightarrow \infty} \left| \frac{k!}{(k+1)!} \right|\] Further to: \[\lim_{k \rightarrow \infty} \left| 1/(k+1) \right|\] Which is clearly 0 since as k becomes larger, the whole expression tends to 0. Hence the series absolutely converges because 0 < 1.
02

Test for Conditional Convergence

Even though we have proved the series to be absolutely convergent, let's proceed with the Alternating Series Test only for the educational purposes. It states that if the terms decrease monotonically to 0, the alternating series converges. i.e., if \(|b_{n+1}| ≤ |b_n|\) for all sufficiently large n and \(\lim_{n \rightarrow \infty} b_n = 0\), then the series converges. Here, each 'b' term is \(3/k!\). As k increases, each consecutive term decreases and approaches 0, hence it confirms that the series is conditionally convergent. However, this step is redundant because if a series is absolutely convergent, it is also conditionally convergent.
03

Final Conclusion

The series \(\sum_{k=0}^{\infty}(-1)^{k} \frac{3}{k !}\) is both absolutely and conditionally convergent. This is determined based on the Ratio Test and the Alternating Series Test.

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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 valuable tool for determining the convergence of an infinite series. By examining the ratio of consecutive terms, we can predict the behavior of the series. For a series \( \sum a_n \), consider the terms \( a_{n+1} \) and \( a_n \). We analyze the ratio \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If this limit is less than 1, the series is absolutely convergent.
  • If the limit equals more than 1, the series diverges.
  • If it equals 1, the test does not give a conclusion.
Breaking down the Ratio Test can reveal insights not initially apparent, especially for factorials like \( \frac{3}{n!} \). The factorial grows very quickly, which can lead the ratio to shrink towards zero when applied, indicating absolute convergence.
Alternating Series Test
The Alternating Series Test is specifically used when a series alternates between positive and negative terms. It provides a method to determine conditional convergence. Consider a series \( \sum (-1)^n b_n \). The conditions for convergence are:
  • The absolute value of the terms \( b_n \) must be decreasing.
  • As \( n \) approaches infinity, the term \( b_n \) should approach zero.
For the series \( \sum (-1)^n \frac{3}{n!} \), these conditions apply because each term reduces in value rapidly and approaches zero. This validation of monotonic decrease coupled with the eventual vanishing of terms ensures the series is conditionally convergent, although here the series is already absolutely convergent.
Factorials
Factorials play a crucial role in series, especially in calculations involving sequences where terms include factorial expressions. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). The factorial \( k! \) grows exceedingly fast, far outweighing polynomial or even exponential growth as \( k \) increases. This rapid growth causes series terms like \( \frac{3}{k!} \) to become extremely small very quickly, which is a catalyst for convergence in many series.
Absolute and Conditional Convergence
The series convergence can be categorized into absolute and conditional.Absolute convergence means that when you take the absolute value of each term in the series, the resulting series converges. A series that is absolutely convergent is also conditionally convergent.Conditional convergence occurs when the series converges with the alternating sign, but diverges when absolute values of the terms are considered. In our exercise with the series \( \sum (-1)^k \frac{3}{k!} \), the absolute convergence was determined using the Ratio Test, since the factorial terms rapidly grow causing the series' terms to reduce in size significantly.Understanding these concepts is key to accurately analyzing infinite series and determining their convergence state.

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Most popular questions from this chapter

You may have wondered why it is necessary to show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) to conclude that a Taylor series converges $$\text { to } f(x) . \text { Consider } f(x)=\left\\{\begin{array}{l} e^{-1 / x^{2}} . \\ 0 \end{array}\right.$$ if \(x \neq 0\) show that if \(x=0\) \(f^{\prime}(0)=f^{\prime \prime}(0)=0 .\) (Hint: Use the fact that \(\lim _{h \rightarrow 0} \frac{e^{-1 / h^{2}}}{h^{n}}=0\) for any positive integer \(n .\) ) It turns out that \(f^{(n)}(0)=0\) for all \(n .\) Thus, the Taylor series of \(f(x)\) about \(c=0\) equals 0 , a convergent "scries" that does not converge to \(f(x)\)

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