Problem 11
Question
Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{k = 1}^{\infty} \frac {1}{k!} \)
Step-by-Step Solution
Verified Answer
The series is convergent by the Ratio Test.
1Step 1: Identify the General Term
In the given series, the general term is \( a_k = \frac{1}{k!} \). This is the term we will use to apply the Ratio Test.
2Step 2: Apply the Ratio Test Formula
The Ratio Test states that for a series \( \sum a_k \), we consider the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). If \( L < 1 \), the series converges; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.
3Step 3: Calculate the Ratio \( \frac{a_{k+1}}{a_k} \)
Calculate \( \frac{a_{k+1}}{a_k} = \frac{1}{(k+1)!} \div \frac{1}{k!} = \frac{k!}{(k+1)!} \). Simplify this to \( \frac{1}{k+1} \).
4Step 4: Take the Limit as \( k \to \infty \)
Now, calculate the limit: \( L = \lim_{k \to \infty} \left| \frac{1}{k+1} \right| = \lim_{k \to \infty} \frac{1}{k+1} = 0 \).
5Step 5: Interpret the Result
Since the limit \( L = 0 \) and \( L < 1 \), by the Ratio Test, the series \( \sum_{k=1}^{\infty} \frac{1}{k!} \) is convergent.
Key Concepts
Convergence and Divergence of SeriesFactorials in SeriesLimit of a Sequence
Convergence and Divergence of Series
Understanding whether a series converges or diverges is a fundamental aspect of series analysis. A series is essentially the sum of the terms of a sequence. When we talk about convergence, we're asking if as we add more and more terms in the series, do we approach a specific value?
If the series approaches a specific number, it is convergent. If it does not, we say that the series is divergent. Let's put this in perspective:
If the series approaches a specific number, it is convergent. If it does not, we say that the series is divergent. Let's put this in perspective:
- Convergent series reach a limit, meaning they will sum to a finite value as more terms are added.
- Divergent series do not approach a limit; they either grow indefinitely or oscillate without settling on a fixed number.
- Analyzing convergence helps us determine the behavior of infinite series, which is useful in calculus and other areas of mathematics.
Factorials in Series
Factorials play a crucial role in series, especially in problems involving sequences with rapidly increasing terms. A factorial, represented as \( k! \), is the product of all positive integers from 1 to \( k \). For example,
- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
- In the series \( \sum_{k=1}^{\infty} \frac{1}{k!} \), the terms decrease rapidly because the denominator, \( k! \), gets large much quicker than an exponential function.
- This rapid decrease in term size contributes to the convergence of many series because the terms shrink quickly enough to sum to a finite value.
Limit of a Sequence
The concept of limits is central to calculus and series analysis. When examining the limit of a sequence, we're looking to see what value a sequence approaches as it continues infinitely.
In the Ratio Test, like in our problem, limits are used to decide on convergence or divergence:
Understanding limits not only helps in series convergence but is also vital in defining derivatives, integrals, and continuity in broader calculus themes.
In the Ratio Test, like in our problem, limits are used to decide on convergence or divergence:
- If the limit \( L \) of the ratio of consecutive terms is less than 1, the series converges.
- If \( L > 1 \), the series diverges.
- When \( L = 1 \), the test does not provide information, and another method is needed.
Understanding limits not only helps in series convergence but is also vital in defining derivatives, integrals, and continuity in broader calculus themes.
Other exercises in this chapter
Problem 11
Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {( - 1)^n 4^n}{\sqrt{n}} x^n \)
View solution Problem 11
Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \left( \frac {1}{n^3} + \frac {1}{3^n} \right) \)
View solution Problem 11
Determine whether the series converges or diverges. \( \displaystyle \sum_{k = 1}^{\infty} \frac { \sqrt [3]{k}}{\sqrt {k^3 + 4k + 3}} \)
View solution Problem 11
Determine whether the series is convergent or divergent. \( 1 + \frac {1}{8} + \frac {1}{27} + \frac {1}{64} + \frac {1}{125} + \cdot \cdot \cdot \)
View solution