/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(A_{n}:=1+\left(1 / 2^{2}\r... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 18

Let \(A_{n}:=1+\left(1 / 2^{2}\right)+\cdots+\left(1 / n^{2}\right)\) for \(n \in \mathbb{N}\). Show that there is no real number \(\alpha<1\) such that \(\left|A_{n+1}-A_{n}\right| \leq \alpha\left|A_{n}-A_{n-1}\right|\) for all \(n \in \mathbb{N}\) with \(n \geq 2\), but \(\left(A_{n}\right)\) is a Cauchy sequence.

Short Answer

Expert verified
To summarize, we showed that there is no real number \(\alpha < 1\) satisfying the inequality \(|A_{n+1} - A_n| \leq \alpha |A_n - A_{n-1}|\) by proving a contradiction when assuming there exists such an \(\alpha\). Additionally, we showed that \((A_n)\) is a Cauchy sequence by demonstrating that for all \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that \(\left|A_m - A_n\right| < \epsilon\) for all \(m, n > N\).

Step by step solution

01

Simplify the given inequality

Let's first rewrite the given inequality by substituting \(A_n\) into it. \[|A_{n+1} - A_n| \leq \alpha |A_n - A_{n-1}|\] \[|1 \div (n+1)^2| \leq \alpha |1 \div n^2 - 1 \div (n-1)^2|\]
02

Show contradiction by assuming there exists an \(\alpha < 1\) satisfying the inequality

Assume that there exists an \(\alpha < 1\) such that the inequality from step 1 holds for all \(n \in \mathbb{N}\) with \(n \geq 2\). Then, it must hold for all \(n \geq 2\): \[1 \div (n+1)^2 \leq \alpha \left(1 \div n^2 - 1 \div (n-1)^2\right)\] Divide both sides of the inequality by \(\alpha\): \[\frac{1}{\alpha} \cdot \frac{1}{(n+1)^2} \leq \frac{1}{n^2} - \frac{1}{(n-1)^2}\] Now let's find an upper bound for the sequence differences: \[\frac{1}{(n+1)^2} \leq \alpha \left(\frac{1}{n^2} - \frac{1}{(n-1)^2}\right) \leq \frac{1}{(n-1)^2} - \frac{1}{n^2}\] Hence, \(\frac{1}{\alpha} \cdot \frac{1}{(n+1)^2} \leq \frac{1}{(n-1)^2} - \frac{1}{n^2}\) for all \(n \geq 2\). Taking the sum for \(n = 2\) to \(n = k\), we get: \[\frac{1}{\alpha} \sum_{n=2}^k \frac{1}{(n+1)^2} \leq \frac{1}{(1)^2} - \frac{1}{(k+1)^2}\] Now, let \(k\) go to infinity: \[\frac{1}{\alpha} \sum_{n=2}^\infty \frac{1}{(n+1)^2} \leq \frac{1}{(1)^2}\] This implies that the sum of the left-hand side converges, which is a contradiction since \(\alpha < 1\). Consequently, there is no real number \(\alpha < 1\) that satisfies the inequality.
03

Show that \(\left(A_n\right)\) is a Cauchy sequence

To show that \(\left(A_n\right)\) is a Cauchy sequence, we need to prove that for any \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that \(\left|A_m - A_n\right| < \epsilon\) for all \(m, n > N\). Consider \(\left|A_m - A_n\right|\). Without loss of generality, we can assume that \(m > n\). Then we have: \[\left|A_m - A_n\right| = \left|\frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{m^2} \right|\] Now, using the comparison test for a bounded sum: \[\left|\frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{m^2} \right| \leq \left|\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)} + \cdots + \frac{1}{m(m-1)}\right|\] Observe that this is a telescoping sum, which simplifies to: \[\left|\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)} + \cdots + \frac{1}{m(m-1)}\right| = \left|\frac{1}{n+1} - \frac{1}{m}\right|\] For any \(\epsilon > 0\), we can find an \(N \in \mathbb{N}\) such that \(\left|\frac{1}{n+1} - \frac{1}{m}\right| < \epsilon\) if \(m, n > N\). This implies that \(\left(A_n\right)\) is a Cauchy sequence. Thus, we have shown that there is no real number \(\alpha < 1\) that satisfies the given inequality, but \((A_n)\) is a Cauchy sequence.

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Most popular questions from this chapter

If \(\left(a_{n}\right)\) is a sequence in \(\mathbb{R}\) such that \(a_{n} \neq 0\) for all \(n, \lim _{n \rightarrow \infty}\left|a_{n+1} / a_{n}\right|\) exists and it is less than 1, then show that \(a_{n} \rightarrow 0\).

Let \(y\) be any real number with \(0 \leq y<1\). Define sequences \(\left(b_{n}\right)\) and \(\left(y_{n}\right)\) iteratively as follows. Let \(y_{1}:=10 y\) and \(b_{1}:=\left[y_{1}\right]\), and for each \(n \in \mathbb{N}\), $$ y_{n+1}:=10\left(y_{n}-b_{n}\right) \quad \text { and } \quad b_{n+1}:=\left[y_{n+1}\right] \text { . } $$ Show that for each \(n \in \mathbb{N}\) we have $$ 0 \leq y_{n}<10 \quad \text { and } \quad b_{n} \in \mathbb{Z} \text { with } 0 \leq b_{n} \leq 9 $$ and moreover, $$ y=\frac{b_{1}}{10}+\frac{b_{2}}{10^{2}}+\cdots+\frac{b_{n}}{10^{n}}+\frac{y_{n+1}}{10^{n+1}} $$ Deduce that $$ 0 \leq \frac{y_{n+1}}{10^{n+1}}<\frac{1}{10^{n}} \quad \text { for each } n \in \mathbb{N} $$ and consequently, $$ y=\lim _{n \rightarrow \infty}\left(\frac{b_{1}}{10}+\frac{b_{2}}{10^{2}}+\cdots+\frac{b_{n}}{10^{n}}\right). $$ [Note: It is customary to call the nonnegative integers \(b_{1}, b_{2}, \ldots\), the digits of \(y\) and write the above expression for \(y\) as \(y=0 . b_{1} b_{2} \ldots\), and call it the decimal expansion of \(y\).]

Prove that a sequence \(\left(a_{n}\right)\) in \(\mathbb{R}\) has no convergent subsequence if and only if \(\left|a_{n}\right| \rightarrow \infty\).

Let \(A_{n}:=1+(1 / 2)+\cdots+(1 / n)\) for \(n \in \mathbb{N}\). Show that \(\left(A_{n+1}-A_{n}\right) \rightarrow 0\) as \(n \rightarrow \infty\), but \(\left(A_{n}\right)\) is not a Cauchy sequence.

A real number \(a\) is called a cluster point of a sequence \(\left(a_{n}\right)\) in \(\mathbb{R}\) if there is a subsequence \(\left(a_{n_{k}}\right)\) of \(\left(a_{n}\right)\) such that \(a_{n_{k}} \rightarrow a\) (i) Show that if \(a_{n} \rightarrow a\), then \(a\) is the only cluster point of \(\left(a_{n}\right)\). (ii) Show that the converse of (i) is not true. In other words, show that there is a divergent sequence that has a unique cluster point. (Hint: \(a_{2 k}:=\frac{1}{2 k}\) and \(a_{2 k+1}:=2 k+1\) for \(k \in \mathbb{N}\).) (iii) Show that if \(a_{n} \rightarrow \infty\) or if \(a_{n} \rightarrow-\infty\), then \(\left(a_{n}\right)\) has no cluster point. (iv) Show that the converse of (iii) is not true. In other words, show that there is a sequence without a cluster point that neither tends to \(\infty\) nor tends to \(-\infty .\) (Hint: \(a_{n}:=(-1)^{n} n\) for \(n \in \mathbb{N}\).)
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