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Chapter 3: Problem 6

Prove that if \(a_{n} \rightarrow+\infty\) and if \(\left\\{\boldsymbol{b}_{n}\right\\}_{n=1}^{+\infty}\) is bounded, then \(a_{n}+b_{n} \rightarrow+\infty\)

Short Answer

Expert verified
If \(a_n \to +\infty\) and \(b_n\) is bounded, then \(a_n + b_n \to +\infty\).

Step by step solution

01

Understand the Given Conditions

We are given that \(a_n \rightarrow +\infty\), which means for every real number \(M\), there exists a natural number \(N\) such that for all \(n > N\), \(a_n > M\). Also, \(\{b_n\}\) is a bounded sequence, meaning there exists some real number \(K\) such that \(|b_n| \leq K\) for all \(n\).
02

Establish the Target Condition

We need to show that the sequence \(a_n + b_n\) tends to \(+\infty\). This means that for every real number \(L\), there exists a natural number \(N'\) such that for all \(n > N'\), \(a_n + b_n > L\).
03

Calculate the Necessary Bound

Since \(b_n\) is bounded by \(K\), we have \(-K \leq b_n \leq K\). Given \(a_n\) tends to \(+\infty\), for any \(L-K\), there exists some \(N\) such that for all \(n > N\), \(a_n > L - K\).
04

Prove the Convergence

For \(n > N\), we have \(a_n > L - K\). Therefore, \(a_n + b_n \geq a_n - K > (L - K) - K = L\). Thus, \(a_n + b_n > L\) for all \(n > N\), proving that \(a_n + b_n \rightarrow +\infty\).

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Key Concepts

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

Bounded Sequence
A bounded sequence is a collection of numbers where all the elements fall within a fixed range. Specifically, a sequence \( \{ b_n \} \) is said to be bounded if there exists a real number \( K \) such that the absolute value of each term in the sequence does not exceed \( K \). Mathematically, this means that \( |b_n| \leq K \) for all terms \( n \) in the sequence.
  • A positive \( K \) ensures that the sequence does not grow indefinitely beyond a certain magnitude.
  • Boundedness provides a constraint which simplifies exploration and manipulation of sequences, particularly when combined with other sequences.
Understanding that a sequence is bounded is crucial in analyzing how it interacts with other sequences. For instance, when combined with an unbounded sequence, such as one that is divergent to infinity, a bounded sequence can reveal important properties about the convergence of their sum.
Real Number Limit
The concept of a real number limit refers to whether the terms of a sequence approach a specific real number as they extend indefinitely. In other words, a sequence \( \{ a_n \} \) is said to have a limit \( L \) if the terms of the sequence become arbitrarily close to \( L \) as \( n \) becomes large.
  • If \( a_n \rightarrow L \), it means that for every small positive number \( \epsilon \), there exists a natural number \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| \) is smaller than \( \epsilon \).
  • This concept is central to understanding and proving the behavior of sequences as they progress, allowing us to predict long-term behavior based on short-term observations.
In the context of unbounded sequences, we consider what happens as their limits approach infinity instead of settling on a finite number.
Natural Number Bound
A natural number bound is a term used in the context of establishing convergence or bounds in sequences. Specifically, when we say that for a given condition \( a_n \rightarrow +\infty \), there exists a natural number \( N \), we mean that starting from some point in the sequence, specifically the \( N \)-th term, all successive terms of the sequence comply with a particular property. For instance, in this problem, the property is that \( a_n > M \) for all \( n > N \).
  • Using a natural number bound allows us to concretely pinpoint where a certain behavior begins in a sequence.
  • This is vital when we need rigorous proof elements, as it precisely identifies beyond which term all conditions hold true.
Thus, a natural number bound can effectively demonstrate how sequences that tend towards infinity remain consistent in their behavior beyond a certain point.
Infinite Limit Proof
Proving an infinite limit involves demonstrating that as you progress through the sequence, the terms exceed any potential finite bound. Specifically, for a sequence \( \{ a_n \} \) that we claim to diverge to infinity, we need to validate that for any large number \( L \), there exists a natural number \( N \) where for all terms beyond \( n > N \), \( a_n > L \).
  • This method of proof is done by isolating a natural number \( N \) which provides the cut-off beyond which all terms meet the stated condition.
  • To establish that \( a_n + b_n \rightarrow +\infty \), knowing \( a_n \) already diverges to infinity and \( b_n \) is bounded, helps consolidate the argument.
The proof then hinges on showing \( a_n + b_n \) surpasses any given real number, leveraging the boundedness of \( \{ b_n \} \) to facilitate \( \{ a_n \} \)'s tendency to infinite growth.

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Most popular questions from this chapter

For a sequence of real numbers \(\left\\{a_{n}\right\\}_{n=1}^{+\infty}\) it is known that \(a_{n}=O\left(n^{2}\right)\) and \(a_{n}=o\left(n^{2}\right) .\) Which of the tuo statements conveys more information?

What's wrong with the following? Since the product of the limits is the limit of the product, $$ e=\lim _{n \rightarrow+\infty}\left(1+\frac{1}{n}\right)^{n}=\underbrace{\left(\lim _{n \rightarrow+\infty} 1+\frac{1}{n}\right) \cdot\left(\lim _{n \rightarrow+\infty} 1+\frac{1}{n}\right) \cdots\left(\lim _{n \rightarrow+\infty} 1+\frac{1}{n}\right.} $$

(Gram's Product) Prove that $$ \lim _{n \rightarrow+\infty} \prod_{k=2}^{n} \frac{k^{3}-1}{k^{3}+1}=\frac{2}{3} $$

Prove that if \(a_{n} \rightarrow+\infty\) then \(\frac{1}{a_{n}} \rightarrow 0\).

Prove that \(\frac{1}{(n !)^{1 / n}} \rightarrow \mathbf{0}\)
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