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Chapter 11: Problem 116

Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\epsilon\) there corresponds an integer \(N\) such that for all \(m\) and \(n\) , $$ m>N \text { and } n>N \Rightarrow\left|a_{m}-a_{n}\right|<\epsilon $$

Short Answer

Expert verified
If the sequence is convergent, there exists an integer \( N \) such that for all \( m, n > N \), \(|a_m - a_n| < \epsilon \).

Step by step solution

01

Identify the Limit

Let's say the sequence \( \{a_n\} \) converges to the limit \( L \). This means we have \( \lim_{{n \to \infty}} a_n = L \).
02

Define Convergent Sequence Condition

By the definition of a convergent sequence, for every positive number \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n > N \), the terms of the sequence satisfy \( |a_n - L| < \frac{\epsilon}{2} \).
03

Consider the Difference Between Terms

Consider two terms of the sequence, \( a_m \) and \( a_n \), where both indices \( m \) and \( n \) are greater than \( N \). We need to show that \( |a_m - a_n| < \epsilon \).
04

Apply Triangle Inequality

Using the triangle inequality, we have: \[ |a_m - a_n| = |a_m - L + L - a_n| \leq |a_m - L| + |a_n - L|. \]
05

Use Convergent Sequence Condition

Since both \( m > N \) and \( n > N \), we have \( |a_m - L| < \frac{\epsilon}{2} \) and \( |a_n - L| < \frac{\epsilon}{2} \). Thus,\[ |a_m - a_n| \leq |a_m - L| + |a_n - L| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]
06

Conclusion

We have shown that for every \( \epsilon > 0 \), there is an integer \( N \) such that for all \( m > N \) and \( n > N \), \( |a_m - a_n| < \epsilon \). This completes the proof that the sequence is Cauchy and therefore the given condition holds.

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Key Concepts

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

Cauchy Sequence
A Cauchy sequence is a powerful concept in mathematics, particularly in the study of sequences and series. It is a type of sequence where the elements get arbitrarily close to each other as the sequence progresses. In simpler terms, a sequence \( \{a_n\} \) is Cauchy if, for every small distance \( \epsilon > 0 \), there exists an index \( N \) such that every pair of terms in the sequence that come after this index are within \( \epsilon \) distance of each other. This can be formally written as:
  • For all \( m, n > N \), \( |a_m - a_n| < \epsilon \).
Think of sailing a boat to a coastline: as you sail, you get closer to the coast, never veering too far away. A key feature of Cauchy sequences is that they promise convergence in complete spaces, like real numbers. This means if a sequence is Cauchy, it likely points to a limit that the sequence is approaching.
Triangle Inequality
The triangle inequality is a fundamental property in mathematics, especially essential in sequence and distance concepts. This inequality states that for any real or complex numbers, the absolute value of the sum is less than or equal to the sum of the absolute values. It can be mathematically described as:
  • \( |x + y| \leq |x| + |y| \)
Imagine walking between three points forming a triangle: the direct path (one side) is always shorter or equal to the sum of lengths when you go through the third point (two sides). In our sequence, applying the triangle inequality:
  • For any terms \( a_m \) and \( a_n \), \( |a_m - a_n| = |a_m - L + L - a_n| \leq |a_m - L| + |a_n - L| \)
This step is crucial because it helps break down the difference between two sequence elements into manageable parts. Ultimately, it shows how close the elements are based on their proximity to the limit \( L \).
Limit of a Sequence
The limit of a sequence is a core concept in calculus and analysis, describing the value that the elements of the sequence approach as the index becomes infinitely large. When we say a sequence \( \{a_n\} \) converges to a limit \( L \), we mean that the terms get closer and closer to \( L \) as \( n \) increases, with their distance to \( L \) becoming arbitrarily small. Formally, it is defined as:
  • For every \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \).
Think of a target that you aim for repeatedly. As you practice, your shots zero in on the bullseye鈥攖he limit \( L \). Achieving a clear visual of the sequence's behavior, the limit gives a precise description of its destination, rewarding every effort to pinpoint its location. Understanding limits brings clarity in solving many problems where the behavior of sequences matters greatly.

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Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}-y=-x, \quad y^{\prime}(2)=-2 \text { and } y(2)=0 $$

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.1} \frac{1}{\sqrt{1+x^{4}}} d x $$

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x) .\) Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) , $$ \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) $$

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)
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