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Chapter 2: Problem 12

Let \(t_{1}=1\) and \(t_{n+1}=\left[1-\frac{1}{(n+1)^{2}}\right] \cdot t_{n}\) for \(n \geq 1\) (a) Show that \(\lim t_{n}\) exists. (b) What do you think lim \(t_{n}\) is? (c) Use induction to show that \(t_{n}=\frac{n+1}{2 n}\) (d) Repeat part (b).

Short Answer

Expert verified
The limit exists and is \( \frac{1}{2} \).

Step by step solution

01

Analyze the Recurrence Relation

Observe the given recurrence relation: \[ t_{n+1} = \left[ 1 - \frac{1}{(n+1)^{2}} \right] t_{n}. \] Identify the base term: \( t_{1} = 1 \).
02

Set Up and Analyze the Limit

Assume \( t_{n} \) approaches a limit \( L \) as \( n \to \infty \). Using the recurrence relation in the limit, \[ L = \left( 1 - \frac{1}{(\text{{large number}})^{2}} \right)L. \] This simplifies to \( L = L \), indicating the limit exists. We will re-evaluate after part (c).
03

Guess the Limit

Intuitively, for very large \( n \), the term \( \frac{1}{(n+1)^{2}} \) approaches 0, and the multiplication factor approaches 1. So \( L \) could be 0.
04

Introduce the Induction Hypothesis

To find an explicit form, use induction. Assume \( t_{k} = \frac{k+1}{2k} \) for some \( k \geq 1 \). Prove that \( t_{k+1} = \frac{(k+1)+1}{2(k+1)} \).
05

Prove the Base Case

For \( t_{1} = 1 \), evaluate: \[ t_{1} = \frac{1+1}{2 \times 1} = 1. \] The base case holds true.
06

Prove the Induction Step

Suppose for some \( n = k \), \( t_{k} = \frac{k+1}{2k} \). Using the recurrence: \[ t_{k+1} = \[ 1 - \frac{1}{(k+1)^{2}} \] \frac{k+1}{2k}. \] Simplify:\[ t_{k+1} = \frac{k+1}{2k} \left(1 - \frac{1}{(k+1)^{2}} \right) = \frac{k+1}{2k} \left( \frac{(k+1)^{2} - 1}{(k+1)^{2}} \right) = \frac{k+1}{2k} \left( \frac{k^{2} + 2k}{(k+1)^{2}} \right) = \frac{(k+1)k}{2k(k+1)} = \frac{k+2}{2(k+1)}. \] Therefore, \( t_{k+1} = \frac{k+2}{2(k+1)} \).
07

Conclude Using Induction

By induction, \( t_{n} = \frac{n+1}{2n} \) holds for all \( n \geq 1 \).
08

Re-evaluate the Limit

Using \( t_{n} = \frac{n+1}{2n} \) and taking the limit as \( n \to \infty \), \[ \lim_{n \to \infty} t_{n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}. \]

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Key Concepts

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

recurrence relations
A recurrence relation defines each term of a sequence using the previous terms. In our exercise, the sequence is defined as follows:
  • Base term: t1 = 1
  • Recurrence relation: tn+1 = (1 - 1/(n+1)2) * tn

To find future terms, we repeatedly apply this relation. Recurrence relations are useful in modeling dynamic processes and finding complex patterns.
mathematical induction
Mathematical induction is a powerful tool to prove statements about all natural numbers. With induction, we follow these steps:
  • Base Case: Check the statement for the initial value.
  • Induction Hypothesis: Assume the statement is true for some k.
  • Induction Step: Prove the statement for k+1 using the hypothesis.

In our exercise, we use induction to prove the sequence formula tn = (n+1)/(2n). Once established, this formula helps determine large terms without repeated calculations.
Here's a quick outline:
  • Base Case: For n=1, t1 = 1::
    • t1 = (1+1)/(2*1) = 1. The base case holds.

  • Induction Step: Assume tk = (k+1)/(2k). Prove tk+1 = (k+2)/(2(k+1)):
    tk+1 = (1 - 1/(k+1)2)(k+1)/(2k)
    • By simplifying we get tk+1 = (k+2)/(2(k+1)). Thus, the induction step holds.

    With both the base case and the induction step verified, the statement is proven for all natural numbers.
    • limit of a sequence
    The limit of a sequence is the value that the terms of a sequence approach as the term index increases indefinitely. For our sequence, tn, we aim to find this limit.

    Observe the recurrence relation:
    tn+1 = (1 - 1/(n+1)2) * tn . Assume the limit exists and denote it by L. Applying the limit: 
    L = (1 - 1/(infinity)2) * L

    Since 1/(infinity)2 tends to 0, simplifying gives L=L. This suggests any value could be a solution.

    Using the explicit formula:
     tn = (n+1)/(2n)

    Taking the limit as n -> infinity:
    L = limn -> infinity (n+1)/(2n) = 1/2
    So, tn approaches 1/2 as n grows very large.
    explicit formula
    An explicit formula gives the general term of a sequence directly, bypassing the need for recurrence relations. For our sequence, the explicit formula derived is tn = (n+1)/(2n). With this, any term can be computed without knowledge of the previous terms.

    Deriving the explicit formula often involves techniques like:
    • Pattern recognition
    • Using mathematical induction
    In our exercise, we initially used induction to show tk = (k+1)/(2k) and then proved it for all n. This formula not only simplifies term computation but also aids in understanding the behavior of the sequence as n becomes large. For instance, the limit calculation becomes straightforward using the explicit formula.
    Explicit formulas help in quickly obtaining terms and analyzing sequence properties without recursive calculations.

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    Most popular questions from this chapter

    Let \(s_{1}=1\) and \(s_{n+1}=\left(\frac{n}{n+1}\right) s_{n}^{2}\) for \(n \geq 1\) (a) Find \(s_{2}, s_{3}\) and \(s_{4}\) (b) Show that \(\lim s_{n}\) exists. (c) Prove that \(\lim s_{n}=0\)

    Let \(s_{1}=1\) and \(s_{n+1}=\frac{1}{3}\left(s_{n}+1\right)\) for \(n \geq 1\) (a) Find \(s_{2}, s_{3}\) and \(s_{4}\) (b) Use induction to show that \(s_{n}>\frac{1}{2}\) for all \(n\) (c) Show that \(\left(s_{n}\right)\) is a nonincreasing sequence. (d) Show that \(\lim s_{n}\) exists and find \(\lim s_{n} .\)

    Suppose that there exists \(N_{0}\) such that \(s_{n} \leq t_{n}\) for all \(n>N_{0}\). (a) Prove that if \(\lim s_{n}=+\infty,\) then \(\lim t_{n}=+\infty\) (b) Prove that if \(\lim t_{n}=-\infty,\) then \(\lim s_{n}=-\infty\) (c) Prove that if \(\lim s_{n}\) and \(\lim t_{n}\) exist, then \(\lim s_{n} \leq \lim t_{n} .\)

    (a) Show that if \(\lim s_{n}=+\infty\) and \(k>0,\) then \(\lim \left(k s_{n}\right)=+\infty\) (b) Show that \(\lim s_{n}=+\infty\) if and only if \(\lim \left(-s_{n}\right)=-\infty\) (c) Show that if \(\lim s_{n}=+\infty\) and \(k<0,\) then \(\lim \left(k s_{n}\right)=-\infty\)

    Let \(s_{1}=1\) and for \(n \geq 1\) let \(s_{n+1}=\sqrt{s_{n}+1}\) (a) List the first four terms of \(\left(s_{n}\right)\) (b) It turns out that \(\left(s_{n}\right)\) converges. Assume this fact and prove that the limit is \(\frac{1}{2}(1+\sqrt{5})\)
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