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Magazine Chapter 9: Problem 56
Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$
Short Answer
Expert verified
Answer: The series converges conditionally.
Step by step solution
01
Determine the Type of Convergence by the Alternating Series Test
To apply the Alternating Series Test, we need to ensure that two conditions are met:
1. The terms of the sequence should decrease in absolute value, i.e., (write the sequence without the (-1) term)
2. The limit of the sequence should equal 0.
Let's call the series $$S$$, then:
$$S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$
Now, let's find the absolute value of the sequence without the (-1) term:
$$a_k = \frac{e^{k}}{(k+1)!}$$
02
Show That the Terms Decrease in Absolute Value
To show that the terms of the sequence $$a_k = \frac{e^{k}}{(k+1)!}$$ decrease, we must show that $$a_{k+1} < a_k$$ for all $$k$$. Let's analyze the ratio of consecutive terms,
$$\frac{a_{k+1}}{a_k} = \frac{\frac{e^{k+1}}{(k+2)!}}{\frac{e^{k}}{(k+1)!}}$$
To find the ratio, we can simplify the expression:
$$\frac{a_{k+1}}{a_k} = \frac{e^{k+1}(k+1)!}{e^k (k+2)!} = \frac{e}{k+2}$$
Since the ratio is less than 1 for all $$k$$, the terms of $$a_k$$ decrease.
03
Find the Limit of the Sequence
We can now find the limit of the sequence $$a_k$$:
$$\lim_{k\to \infty} \frac{e^{k}}{(k+1) !}$$
Here, we have exponentials in the numerator and factorials in the denominator. We know that factorials grow faster than exponentials, so the limit is:
$$\lim_{k\to \infty} \frac{e^{k}}{(k+1) !} = 0$$
The limit equals 0, so the second condition for the Alternating Series Test is met.
04
Determine Convergence Using the Alternating Series Test
Since both conditions for the Alternating Series Test are met, the original series S converges conditionally. It means that:
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$ converges conditionally.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence of Series
The convergence of a series refers to whether the sum of its infinite terms approaches a finite number as the number of terms increases. In mathematics, not all series will converge; some will diverge, meaning the series sums to infinity. In this case, we examine whether the series converges absolutely, converges conditionally, or diverges.
When using the Alternating Series Test, we identify convergence in an alternating series by checking two key conditions:
- The absolute value of the terms must form a decreasing sequence. Essentially, each term in the series should be smaller in absolute value than the term before it.
- The limit of the sequence of terms must be zero as the number of terms tends to infinity.
Factorials vs Exponentials
Factorials, denoted as \( n! \), represent the product of all positive integers up to \( n \). They grow very rapidly because each number multiplies a larger sequence of integers. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Compare this to an exponential function, \( e^n \), which grows quickly but not as fast as a factorial.In the context of the series \( a_k = \frac{e^k}{(k+1)!} \), we see an interplay between exponential growth in the numerator and factorial growth in the denominator. As \( k \) increases, \( (k+1)! \), the factorial term, increases much faster compared to \( e^k \), the exponential term. This rapid growth of the factorial term effectively diminishes each term in the series more quickly, creating the decrease necessary for convergence as suggested in the original exercise. Thus, when comparing factorials versus exponentials in series, the dominance of the factorial often plays a crucial role in determining whether a series converges or diverges.
Conditional Convergence
Conditional convergence occurs when a series converges according to the Alternating Series Test, but does not converge absolutely. A series converges absolutely if the series of the absolute values of its terms also converges. For our series, after removing the alternating terms \( (-1)^{k+1} \), the resulting series \( \sum_{k=1}^{\infty} \frac{e^k}{(k+1)!} \) does not converge. Since the series \( \sum_{k=1}^{\infty} \frac{e^k}{(k+1)!} \) diverges, but the original series with the alternating sign meets the criteria of the Alternating Series Test, it implies conditional convergence. This simply means that the convergence depends heavily on the alternating sign and the balance it provides.Understanding the concept of conditional convergence is vital, as it reflects a situation where different tests for convergence might yield different insights. It highlights the necessity to carefully scrutinize the absolute convergence alongside conditional convergence to gain a full picture of a series' behavior.
