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Chapter 9: Problem 98

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}.$$

Short Answer

Expert verified
Answer: The limit of the sequence is 0.

Step by step solution

01

Write down the sequence and the limit we want to find

We are asked to find the limit of the sequence defined as follows: $$ a_n = \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{n}\right), $$ for \(n = 2, 3, 4, ...\). The limit we want to find is: $$ \lim_{n \to \infty} a_n $$
02

Simplify the sequence

By factoring out each term, we can write the sequence \(a_n\) as $$ a_n = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot ... \cdot \frac{n-1}{n}. $$
03

Cancel out the common factors

Notice that each term in the sequence has a common factor with its neighboring term, like \(\frac{1}{2}\) with \(\frac{2}{3}\), \(\frac{2}{3}\) with \(\frac{3}{4}\), and so on. Cancel out these common factors to get $$ a_n = \frac{1}{n}. $$
04

Find the limit of the simplified sequence

Now that we have a simplified expression for \(a_n\), finding the limit as \(n\) approaches infinity is straightforward: $$ \lim_{n \to \infty} \frac{1}{n} = 0. $$ Hence, the limit of the sequence is 0.

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Key Concepts

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

Calculus
Calculus is the branch of mathematics that allows us to understand changes and the behavior of functions over time. In our exercise, we're interested in finding the limit of a sequence, which is a common task in calculus. Calculus provides us the tools to deal with sequences and series effectively. The sequence given in the exercise involves products of fractions which change as more terms are added. With calculus, we examine the behavior of this infinite sequence as the number of terms goes to infinity. This process is called taking the limit. By analyzing limits, calculus helps us determine what happens as the input values grow larger and larger. The fundamental theorem of calculus and the techniques for differentiation and integration aren’t directly involved in this specific limit problem, but they anchor the broader field. Being familiar with calculus concepts aids significantly in understanding limits, continuity, and the gradual changes within functions.
Mathematical Sequences
A mathematical sequence is simply a list of numbers in a specific order. Each number in the sequence is termed a 'term.' Sequences can be finite or infinite, and they are defined based on rules or formulas.In our exercise, the sequence is defined as a product of fractions, with each term built on the previous ones:
  • The first term is simple, such as \(1-\frac{1}{2}\).
  • More terms are multiplied subsequently, like \(\left(1-\frac{1}{3}\right)\), forming a sequence from these multiplying fractions.
Mathematical sequences often create intriguing patterns. Understanding sequences often involves simplification, as we've seen with the cancellation of terms, reducing the original complex form to a simpler one, such as \(\frac{1}{n}\). This simplification is crucial because it makes it easier to eventually find limits or other properties of the sequence.
Convergent Series
A sequence like the one given in the exercise, where we find that its terms approach a single value as the sequence progresses, is known as a convergent sequence. When a sequence converges, the terms get closer and closer to a specific value or number.Convergence is a central concept in calculus and is integral for understanding series. A convergent series is a sequence whose terms add up to a specific finite limit as the number of terms becomes infinitely large. In the exercise, the simplification of the sequence presents us with \( \frac{1}{n} \), which is a sequence converging to zero.The sequence is known to converge because its terms, which are progressively smaller, approach zero. Thus, the limit of the sequence as \( n \) approaches infinity is zero. Recognizing a convergent series or sequence is vital for determining the behavior of functions and systems described by these mathematical tools.

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Most popular questions from this chapter

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

The expression where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+1 / a_{n},\) for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of where \(a\) and \(b\) are positive real numbers.

Determine whether the following statements are true and give an explanation or counterexample. a. If \(\lim _{n \rightarrow \infty} a_{n}=1\) and \(\lim _{n \rightarrow \infty} b_{n}=3,\) then \(\lim _{n \rightarrow \infty} \frac{b_{n}}{a_{n}}=3\). b. If \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\lim _{n \rightarrow \infty} b_{n}=\infty,\) then \(\lim _{n \rightarrow \infty} a_{n} b_{n}=0\). c. The convergent sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) differ in their first 100 terms, but \(a_{n}=b_{n},\) for \(n>100 .\) It follows that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}\). d. If \(\left\\{a_{n}\right\\}=\left\\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\right\\}\) and \(\left\\{b_{n}\right\\}=\) \(\left\\{1,0, \frac{1}{2}, 0, \frac{1}{3}, 0, \frac{1}{4}, 0, \ldots\right\\}, \text { then } \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}.\) e. If the sequence \(\left\\{a_{n}\right\\}\) converges, then the sequence \(\left\\{(-1)^{n} a_{n}\right\\}\) converges. f. If the sequence \(\left\\{a_{n}\right\\}\) diverges, then the sequence \(\left\\{0.000001 a_{n}\right\\}\) diverges.

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4} .\) Although you do not need it, the exact value of the series is given in each case. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$
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