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Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ a_{n}=1 / n ! $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges absolutely by the ratio test.

Step by step solution

01

State the Ratio Test

The ratio test states that for a series \( \sum_{n=1}^{\infty} a_n \), if the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) exists, then: 1. If \( L < 1 \), the series converges absolutely.2. If \( L > 1 \), the series diverges.3. If \( L = 1 \), the test is inconclusive.
02

Write the Formula for Ratio

For the given series \( \sum_{n=1}^{\infty} \frac{1}{n!} \), the general term \( a_n \) is \( \frac{1}{n!} \). We need to find \( \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \).
03

Simplify the Ratio

Simplify \( \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \) to get \( \frac{1}{(n+1)!} \cdot n! = \frac{n!}{(n+1)!} = \frac{1}{n+1} \).
04

Compute the Limit

Compute the limit of \( \frac{1}{n+1} \) as \( n \rightarrow \infty \). This is \( \lim_{n \rightarrow \infty} \frac{1}{n+1} = 0 \).
05

Interpret the Result

Since the limit \( L = 0 < 1 \), it follows from the ratio test that the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges absolutely.

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Key Concepts

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

Series Convergence
Series convergence is a fundamental topic in calculus that deals with determining whether the sum of an infinite sequence of terms results in a finite value. An infinite series is written as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the sequence's general term. Convergence implies that as you keep adding terms indefinitely, the total approaches a specific finite number. When considering whether a series converges or diverges, one can employ various tests, such as the Ratio Test, which is particularly handy for series with factorial terms or exponential functions.
Understanding convergence versus divergence is crucial because it informs us how certain infinite processes behave\. If a series converges, we can potentially calculate its sum and use it in practical applications, like computing probabilities or solving differential equations. However, a diverging series grows indefinitely or fluctuates without settling, rendering it often unusable for finding sums and requiring different approaches for analysis.
  • Use tests like the ratio or root tests to determine convergence for complex series.
  • Convergent series simplify analysis in physics and engineering, offering finite answers from infinite sums.
  • Insight into convergence/d divergence is central to the study of calculus and analysis.
Factorial Series
Factorial series are a special type of series where each term involves a factorial, such as \( \frac{1}{n!} \). A factorial is denoted by \( n! \) and is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow very quickly, meaning that the terms \( \frac{1}{n!} \) reduce in size rapidly as \( n \) increases, often leading to convergence when forming a series.
Factorials are often encountered in permutations, combinations, and other combinatorial contexts, making them relevant in statistics and probability. In the context of series, because factorials grow at an exponential pace, they usually help diminish each term's contribution to the sum, making it easier for the series to converge. For instance, the exponential series \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \) converges for any finite value of \( x \), illustrating the power of factorials in providing convergence.
  • Factorial terms shrink quickly, contributing to convergence in many series.
  • Key in solving problems involving permutations and combinations.
  • Vital in applications across mathematics and science, especially in defining ex.
Limit Evaluation
Limit evaluation is an essential skill for determining the behavior of sequences and series as they approach infinity. It often involves finding the limit of a sequence of terms \( a_n \) as \( n \rightarrow \infty \). In the context of the Ratio Test, limit evaluation helps us figure out if a series will converge by finding the value of \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges absolutely.
Understanding how to calculate limits allows us to grasp how terms behave individually and collectively as they infinitely progress. For instance, in our problem, simplifying \( \frac{1}{n+1} \) to evaluate its limit as \( n \) becomes very large shows us that the terms become negligible, which is why the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges. Effective limit evaluation is critical in calculus because it forms the foundation upon which many convergence tests and techniques are built.
  • Key for analyzing the behavior of sequences and series in calculus.
  • Crucial for determining series convergence through tests like the ratio test.
  • Helps predict long-term behavior of a sequence, essential for deeper mathematical insights.

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Most popular questions from this chapter

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.\(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\cdots\)

In the following exercises, use an appropriate test to determine whether the series converges. $$ \left.a_{k}=\left(\frac{k}{k+\ln k}\right)^{2 k} \text { (Hint: } a_{k}=\left(1+\frac{\ln k}{k}\right)^{-(k / \ln k) \ln k^{2}} \cdot\right) $$

The following series converge by the ratio test. Use summation by parts, \(\sum_{k=1}^{n} a_{k}\left(b_{k+1}-b_{k}\right)=\left[a_{n+1} b_{n+1}-a_{1} b_{1}\right]-\sum_{k=1}^{n} b_{k+1}\left(a_{k+1}-a_{k}\right)\), to find the sum of the given series. $$ \left.\sum_{k=1}^{\infty} \frac{k}{c^{k}}, \text { where } c>1 \text { (Hint: Take } a_{k}=k \text { and } b_{k}=c^{1-k} /(c-1) .\right) $$

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.Sometimes the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1} b_{n}\) converges to a certain fraction of an absolutely convergent series \(\sum_{n=1}^{\infty} b_{n}\) a faster rate. Given that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}\), find \(S=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\cdots .\) Which of the series \(6 \sum_{n=1}^{\infty} \frac{1}{n^{2}}\) and \(S \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}\) gives a better estimation of \(\pi^{2}\) using 1000 terms?

The kth term of each of the following series has a factor \(x^{k}\). Find the range of \(x\) for which the ratio test implies that the series converges. $$ \sum_{k=1}^{\infty} \frac{x^{k}}{k !} $$

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