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Chapter 2: Problem 8

Let \(\left(s_{n}\right)\) be a nondecreasing sequence of positive numbers and define \(\sigma_{n}=\frac{1}{n}\left(s_{1}+s_{2}+\cdots+s_{n}\right) .\) Prove that \(\left(\sigma_{n}\right)\) is a nondecreasing sequence.

Short Answer

Expert verified
The sequence \(\sigma_n\) is nondecreasing because \(\sigma_{n+1} - \sigma_n\) is nonnegative.

Step by step solution

01

- Understand the Sequence Definitions

Given a nondecreasing sequence of positive numbers \(s_{n}\), this means for all \(n\), \(s_{n} \leq s_{n+1}\). The sequence \(\sigma_n\) is defined as \(\sigma_{n} = \frac{1}{n}\left(s_{1}+s_{2}+\cdots+s_{n}\right)\).
02

- Express \(\sigma_{n+1}\)

We need to express the next term of the sequence \(\sigma_n\). \(\sigma_{n+1} = \frac{1}{n+1}\left(s_{1}+s_{2}+\cdots+s_{n}+s_{n+1}\right)\).
03

- Compare \(\sigma_{n}\) and \(\sigma_{n+1}\)

Compare \(\sigma_n\) and \(\sigma_{n+1}\) by investigating their difference. \[\sigma_{n+1} - \sigma_n = \frac{1}{n+1}\left(s_{1}+s_{2}+\cdots+s_{n}+s_{n+1}\right) - \frac{1}{n}\left(s_{1}+s_{2}+\cdots+s_{n}\right)\] First, express both terms with a common denominator \(n(n+1)\).
04

- Simplify the Difference

Let鈥檚 simplify the difference step-by-step. \[\frac{1}{n+1}\left(s_{1}+s_{2}+\cdots+s_{n}+s_{n+1}\right) - \frac{1}{n}\left(s_{1}+s_{2}+\cdots+s_{n}\right)\] can be rewritten as \[\frac{n*(s_{1}+s_{2}+\cdots+s_{n}+s_{n+1}) - (n+1)*(s_{1}+s_{2}+\cdots+s_{n})}{n(n+1)}\].
05

- Final Simplification

Simplify the numerator: \[n(s_{1}+s_{2}+\cdots+s_{n}) + ns_{n+1} - n(s_{1}+s_{2}+\cdots+s_{n}) - (s_{1}+s_{2}+\cdots+s_{n}) \] This simplifies to \[\frac{ns_{n+1} - (s_{1}+s_{2}+\cdots+s_{n})}{n(n+1)}\]. Since \(s_{n}\) is nondecreasing, the term \(s_{n+1}\) is at least \(s_{n}\) and thus the numerator is always nonnegative.
06

- Conclusion

Since the numerator is nonnegative, the difference \(\sigma_{n+1} - \sigma_n \geq 0\), proving that the sequence \(\sigma_n\) is nondecreasing.

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Key Concepts

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

Sequence Definitions
To start, let's clarify what a sequence is. In mathematics, a sequence is an ordered list of numbers. For example, \((1, 2, 3, 4, ...)\) is a sequence where each number follows a logical order.
A nondecreasing sequence is a specific type where each term is not less than the one before it. In other words, for a nondecreasing sequence \((s_n)\), we have \((s_n \leq s_{n+1})\) for all \(n)\).
This means the sequence either stays the same or increases as it progresses. Understanding this is critical for working with sequences like the one in our exercise.
Sequence Comparison
To show that one sequence is nondecreasing, we often compare its terms. In the given exercise, we compare terms of the sequence \((\sigma_n)\).
The term \((\sigma_n)\) is defined as \[\sigma_n = \frac{1}{n}(s_1 + s_2 + \cdots + s_n)\] where each \((s_n)\) is a positive number in a nondecreasing sequence.
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.
Average of Sequences
The concept of taking an average is essential in this problem. When we define \((\sigma_n)\) as \[\sigma_n = \frac{1}{n}(s_1 + s_2 + \cdots + s_n)\], we are simply averaging the first \((n)\) terms of \((s_n)\).
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.
This average smooths out any fluctuations, making it easier to analyze the overall trend of the sequence, for example.
Nonnegative Difference
Finally, the idea of a nonnegative difference is key to showing a sequence \(()\) is nondecreasing. In our solution, we showed that \((\sigma_{n+1} - \sigma_n)\) was \((\ \geq 0)\).
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.
Since every term in \((s_n)\) satisfies \((s_n \leq s_{n+1})\), this guarantees that the average \((\sigma_n)\) will either stay the same or increase, making it a nondecreasing sequence.

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Most popular questions from this chapter

(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 \(\left(s_{n}\right)\) be a sequence that converges. (a) Show that if \(s_{n} \geq a\) for all but finitely many \(n,\) then \(\lim s_{n} \geq a.\) (b) Show that if \(s_{n} \leq b\) for all but finitely many \(n,\) then \(\lim s_{n} \leq b.\) (c) Conclude that if all but finitely many \(s_{n}\) belong to \([a, b],\) then \(\lim s_{n}\) belongs to \([a, b].\)

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 \(\lim s_{n} .\)

Let \(\left(s_{n}\right)\) be a sequence of nonnegative numbers, and for each \(n\) define \(\sigma_{n}=\frac{1}{n}\left(s_{1}+s_{2}+\cdots+s_{n}\right)\). (a) Show that $$\lim \inf s_{n} \leq \lim \inf \sigma_{n} \leq \lim \sup \sigma_{n} \leq \lim \sup s_{n}$$ Hint: For the last inequality, show first that \(M>N\) implies \(\sup \left\\{\sigma_{n}: n>M\right\\} \leq \frac{1}{M}\left(s_{1}+s_{2}+\cdots+s_{N}\right)+\sup \left\\{s_{n}: n>N\right\\}\). (b) Show that if \(\lim s_{n}\) exists, then \(\lim \sigma_{n}\) exists and \(\lim \sigma_{n}=\) \(\lim s_{n}\).

We have seen that it is often a lot harder to find the value of an infinite sum than to show that it exists. Here are some sums that can be handled. $$\text { (a) Calculate } \sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n} \text { and } \sum_{n=1}^{\infty}\left(-\frac{2}{3}\right)^{n}$$ $$\begin{aligned} &\text { (b) Prove } \sum_{n=1}^{\infty} \frac{1}{n(n+1)}=1 . \text { Hint: Note that } \sum_{k=1}^{n} \frac{1}{k(k+1)}=\\\ &\sum_{k=1}^{n}\left[\frac{1}{k}-\frac{1}{k+1}\right] \end{aligned}$$ $$\text { (c) Prove that } \sum_{n=1}^{\infty} \frac{n-1}{2^{n+1}}=\frac{1}{2} . \text { Hint: Note that } \frac{k-1}{2^{k+1}}=\frac{k}{2^{k}}-\frac{k+1}{2^{k+1}}$$ (d) Use (c) to calculate \(\sum_{n=1}^{\infty} \frac{n}{2^{n}}\)
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