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Prove that there is a sequence of positive real numbers \(a_{0}, a_{1}, a_{2}, \ldots\) such that the partial sum \(a_{0}+a_{1}+a_{2}+\cdots+a_{n}\) is strictly less than 2 for all \(n \in \mathbb{N}\). Hint: think about how you could define what \(a_{k+1}\) is to make the induction argument work.

Short Answer

Expert verified
By constructing the sequence with \(a_0 = 1\) and \(a_{k+1} = \frac{1}{2}a_k\), and using induction, we prove that the partial sum \(a_0 + a_1 + a_2 + ... + a_n\) is always strictly less than 2 for all \(n \in \mathbb{N}\).

Step by step solution

01

- Understanding the Problem

We want to show that there exists a sequence of positive real numbers where the sum of the first 'n' terms is strictly less than 2 for all natural numbers 'n'. This will involve constructing such a sequence and proving it satisfies the given condition.
02

- Constructing the Sequence

We can define the sequence recursively. Let's start with an initial term, say, \(a_0 = 1\), which is less than 2. Then for each subsequent term, we can define it as \(a_{k+1} = \frac{1}{2}a_k\) with \(k\) being a non-negative integer. This will ensure that each term is positive and each term is half the size of the previous term.
03

- Establishing the Induction Hypothesis

Assume that our induction hypothesis is true for some integer \(n\), such that the sum \(S_n = a_0 + a_1 + ... + a_n < 2\). We'll use induction to prove that this is true for all natural numbers.
04

- The Induction Step

We need to show that if the induction hypothesis holds for \(n\), it also holds for \(n+1\). We have \(S_{n+1} = S_n + a_{n+1}\). By the definition of our sequence, \(a_{n+1} = \frac{1}{2}a_n\) and \(S_n < 2\), hence \(S_{n+1} = S_n + \frac{1}{2}a_n < 2 + \frac{1}{2}a_n\). Since \(a_n < 2 - S_{n-1}\), then \(S_{n+1} < 2 + \frac{1}{2}(2 - S_{n-1}) = 2 + 1 - \frac{1}{2}S_{n-1} < 3 - \frac{1}{2}S_{n-1}\), and clearly \(3 - \frac{1}{2}S_{n-1} < 2\) as \(S_{n-1} < 2\) by induction hypothesis.
05

- Conclusion

Since both our base case and the induction step have been proven, by the principle of mathematical induction, we have shown that the series \(a_0, a_1, a_2, \ldots\) satisfies the condition \(a_0 + a_1 + a_2 + \ldots + a_n < 2\) for all \(n \in \mathbb{N}\).

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Key Concepts

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

Sequence of Real Numbers
A sequence of real numbers is a list of numbers in a specific order, usually denoted as \(a_n\), where \(n\) represents the position of each number in the sequence and \(n \in \mathbb{N}\), the set of natural numbers. These numbers are called the terms of the sequence. Sequences can be finite or infinite, and they are a fundamental concept in mathematics because they can describe patterns and show how quantities evolve over time. For instance, in our exercise, we are looking for an infinite sequence of positive real numbers where each cumulative sum up to any term is strictly less than 2. When conceptualizing such a sequence, we can use patterns or formulas to define the relationship between the terms.
Partial Sum of Series
The partial sum of a series is the sum of the first n terms of an infinite sequence. It is notated as \(S_n = a_0 + a_1 + ... + a_n\). The behavior of these partial sums indicates whether a series converges or diverges, and to what value it might converge if it is convergent. The problem we're tackling involves creating a series where the partial sums never exceed a certain real number, in this case, 2. This means we aim to design a sequence where, no matter how many terms you add together, the sum will always be less than 2; hence, the series is convergent.
Recursive Sequence Definition
A recursive sequence is defined by a starting value and a rule that relates each term to preceding ones. The recursion allows us to generate the rest of the sequence from one or more given initial terms. In our solution, the sequence is defined to start with \(a_0 = 1\), and each subsequent term is half the previous term (\(a_{k+1} = \frac{1}{2}a_k\)). This recursive formula ensures each term is positive, and is instrumental in proving that the sequence's partial sums remain bounded under 2. Such definitions are crucial in mathematical induction, where the proof hinges on showing that if a property holds for a certain case, it will hold for the next case derived from the recursive relationship.

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Most popular questions from this chapter

Consider bit strings with length \(l\) and weight \(k\) (so strings of \(l 0^{\prime} \mathrm{s}\) and \(1^{\prime}\) s, including \(k 1^{\prime}\) s). We know how to count the number of these for a fixed \(l\) and \(k .\) Now, we will count the number of strings for which the sum of the length and the weight is fixed. For example, let's count all the bit strings for which \(l+k=11\). (a) Find examples of these strings of different lengths. What is the longest string possible? What is the shortest? (b) How many strings are there of each of these lengths. Use this to count the total number of strings (with sum 11 ). (c) The other approach: Let \(n=l+k\) vary. How many strings have sum \(n=1\) ? How many have sum \(n=2 ?\) And so on. Find and explain a recurrence relation for the sequence \(\left(a_{n}\right)\) which gives the number of strings with sum \(n\). (d) Describe what you have found above in terms of Pascal's Triangle. What pattern have you discovered?

Consider the sequence \(2,7,15,26,40,57, \ldots\) (with \(a_{0}=2\) ). By looking at the differences between terms, express the sequence as a sequence of partial sums. Then find a closed formula for the sequence by computing the \(n\) th partial sum.

Let \(t_{n}\) denote the number of ways to tile a \(2 \times n\) chessboard using \(1 \times 2\) dominoes. Write out the first few terms of the sequence \(\left(t_{n}\right)_{n \geq 1}\) and then give a recursive definition. Explain why your recursive formula is correct.

Consider the three sequences below. For each, find a recursive definition. How are these sequences related? (a) \(2,4,6,10,16,26,42, \ldots .\) (b) \(5,6,11,17,28,45,73, \ldots\) (c) \(0,0,0,0,0,0,0, \ldots\)

A ternary string is a sequence of \(0^{\prime} \mathrm{s}, 1^{\prime} \mathrm{s}\) and \(2^{\prime} \mathrm{s} .\) Just like a bit string, but with three symbols. Let's call a ternary string good provided it never contains a 2 followed immediately by a \(0 .\) Let \(G_{n}\) be the number of good strings of length \(n .\) For example, \(G_{1}=3,\) and \(G_{2}=8\) (since of the 9 ternary strings of length \(2,\) only one is not good). Find, with justification, a recursive formula for \(G_{n},\) and use it to compute \(G_{5}\)

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