Question

(a) Prove: If \(\left\\{a_{n} r^{n}\right\\}\) is bounded and \(\left|x_{1}-x_{0}\right|R\), then \(\left\\{a_{n}\left(x_{1}-x_{0}\right)^{n}\right\\}\) is unbounded.

Short answer

Short Answer: (a) Given that \(\left\\{a_{n} r^{n}\right\\}\) is bounded, and \(\left|x_{1}-x_{0}\right|R\), we assume by contradiction that the sequence \(\left\\{a_{n}\left(x_{1}-x_{0}\right)^{n}\right\\}\) is bounded. By analyzing the inequalities and the limit superior, we find a contradiction, implying that the sequence must be unbounded.

Step-by-step solution

Definitions of bounded sequences and convergence

A sequence \(\left\\{a_{n}\right\\}\) is said to be bounded if there exists a constant \(M>0\) such that \(|a_{n}| \leq M\) for all \(n\). A series \(\sum a_{n}\) is said to converge if the partial sums \(S_{N}=\sum_{n=0}^{N}a_{n}\) approach a finite limit as \(N \rightarrow \infty\).

Using the boundedness of \(\left\\{a_{n} r^{n}\right\\}\)

Given that \(\left\\{a_{n} r^{n}\right\\}\) is bounded, there exists a constant \(M>0\) such that \(|a_{n} r^n| \leq M\) for all \(n\). Since \(\left|x_{1}-x_{0}\right|<r\), we have \(\frac{\left|x_{1}-x_{0}\right|}{r}<1\). Let \(c=\frac{\left|x_{1}-x_{0}\right|}{r}\).

Comparison Test

Consider the convergent geometric series \(\sum M c^{n}\). We have the inequality \(|a_{n}\left(x_{1}-x_{0}\right)^{n}|=|a_{n}|\cdot{\left|\frac{x_{1}-x_{0}}{r}\right|}^n=|a_{n}r^n|\cdot c^n \leq M c^n\), which implies that \(a_{n}\left(x_{1}-x_{0}\right)^{n}\) is absolutely convergent and \(\sum a_{n}\left(x_{1}-x_{0}\right)^{n}\) converges. This completes the proof for part (a). (b): Prove that the sequence is unbounded

Definition of radius of convergence

The radius of convergence \(R\) is defined as \(R = \frac{1}{\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|}}\).

Assume by contradiction that the sequence is bounded

Suppose, for the sake of contradiction, that \(\left\\{a_{n}\left(x_{1}-x_{0}\right)^{n}\right\\}\) is bounded. Then there exists a constant \(M>0\) such that \(|a_{n}\left(x_{1}-x_{0}\right)^{n}| \leq M\) for all \(n\).

Function outside the radius of convergence

We are given that \(\left|x_{1}-x_{0}\right|>R\). Since \(R = \frac{1}{\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|}}\), we have \(\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|}<\frac{1}{\left|x_{1}-x_{0}\right|}\).

Derive a contradiction

Using the assumption that the sequence is bounded, we have the inequality \(|a_{n}| \leq M\left|x_{1}-x_{0}\right|^{-n}=M\left(\frac{1}{\left|x_{1}-x_{0}\right|}\right)^n\). Taking the nth root of both sides and then the limit superior, we get \(\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|} \leq \frac{1}{\left|x_{1}-x_{0}\right|}\). This contradicts the result from Step 3 that \(\limsup_{n \rightarrow \infty}\sqrt[n]{|a_n|}<\frac{1}{\left|x_{1}-x_{0}\right|}\).

Conclude that the sequence is unbounded

Since our assumption led to a contradiction, we conclude that the sequence \(\left\\{a_{n}\left(x_{1}-x_{0}\right)^{n}\right\\}\) is unbounded. This completes the proof for part (b).

Key concepts

Bounded Sequences

Understanding bounded sequences is crucial when studying the convergence of series. A sequence \( \{a_n\} \) is described as bounded if its terms don't grow indefinitely but instead remain within a certain range, limited by a specific value. Simply put, there's a cap, an upper limit that the terms of the series cannot surpass. Mathematically, this means there exists some constant \( M > 0 \) such that \( |a_n| \leq M \) for all \( n \). This concept can be likened to a fence around a garden, where the plants (the terms of the sequence) cannot grow beyond the fence (the boundary \( M \) ).
When analyzing convergence, boundedness is a property that can guide us toward understanding whether a series has a sum approaching a finite limit. If a sequence is not bounded—picture plants growing uncontrollably with no fence in sight—then we could suspect that the series might diverge, meaning it does not sum up to a specific limit.

Radius of Convergence

The radius of convergence is a measure that indicates the range of input values for which a power series converges. Imagine drawing a circle around a point on a number line; only the values inside this circle will ensure the series converges. Mathematically, if we have a power series \( \sum a_n(x - x_0)^n \) centered at \( x_0 \) with radius of convergence \( R \), it will converge for every \( x \) such that \( |x - x_0| < R \) and diverge for every \( x \) with \( |x - x_0| > R \).
To understand and find the radius of convergence, we often use the ratio or root tests that relate to the terms of the series. In the context of convergence, if you're within this 'magic' circle, the series will nicely add up to a finite sum. Step out of bounds, and the series just won't cooperate—it will diverge, refusing to settle down to a particular value.

Comparison Test

The comparison test is a handy tool in our mathematical toolkit, functioning much like a benchmarking process. When we compare one thing to another to gauge its quality, we use the comparison test to determine whether a series converges or diverges by comparing it to a series whose convergence we already understand. If our 'test series' converges and the terms of the series we're examining are lesser or equal to the corresponding terms of the test series, then our original series will also converge.
This test can be thought of like comparing weights—if a scale can hold the weight of a known heavy object, it can surely hold a lighter one. To perform the comparison test, we first need a series with positive terms that we know converges (like our benchmark), and then we show that the terms of the series we're investigating are not greater than the corresponding terms of our known converging series. If that's the case, we're in luck—our series converges, too!