Question

1.7 We show that the sequence \(\left\\{x_{n}\right\\}_{n=1}^{\infty}\), where \(x_{n}=\sum_{k=1}^{n}(-1)^{k+1} / k\), is Cauchy. Without loss of generality, assume that \(n>m\) and \(n-m\) is even (the case of odd \(n-m\) can be handled similarly). (a) Show that $$ x_{n}-x_{m}=(-1)^{m} \sum_{j=1}^{n-m} \frac{(-1)^{j}}{j+m} . $$ (b) Separate the even and odd parts of the sum and show that $$ x_{n}-x_{m}=(-1)^{m}\left\\{-\sum_{k=1}^{(n-m) / 2} \frac{1}{2 k-1+m}+\sum_{k=1}^{(n-m) / 2} \frac{1}{2 k+m}\right\\} \text { . } $$ (c) Add the two sums to obtain a single sum, showing that $$ x_{n}-x_{m}=-(-1)^{m}\left\\{\sum_{k=1}^{(n-m) / 2} \frac{1}{(2 k+m)(2 k+m-1)}\right\\}, $$ and that $$ \begin{aligned} \left|x_{n}-x_{m}\right| & \leq \sum_{k=1}^{(n-m) / 2} \frac{1}{(2 k+m-1)^{2}} \\\ &=\frac{1}{(1+m)^{2}}+\sum_{k=2}^{(n-m) / 2} \frac{1}{(2 k+m-1)^{2}} . \end{aligned} $$ (d) Convince yourself that \(\int_{1}^{s} f(x) d x \geq \sum_{k=2}^{s} f(k)\) for any continuous function \(f(x)\), and apply it to part (c) to get $$ \begin{aligned} \left|x_{n}-x_{m}\right| & \leq \frac{1}{(1+m)^{2}}+\int_{1}^{(n-m) / 2} \frac{1}{(2 x+m-1)^{2}} d x \\ &=\frac{1}{n}+\frac{1}{(1+m)^{2}}-\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{m+1}\right) . \end{aligned} $$ Each term on the last line goes to zero independently as \(m\) and \(n\) go to infinity.

Short answer

We have proved the sequence to be Cauchy by separating the sum into its even and odd components, combining them into a single sum, and then relating this sum to properties of continuous functions. By proving that the terms go to zero independently as \(m\) and \(n\) go to infinity, we concluded that the sequence satisfies the Cauchy condition.

Step-by-step solution

Defining x_n - x_m

First, we are to prove that \(x_{n}-x_{m}=(-1)^{m} \sum_{j=1}^{n-m} \frac{(-1)^{j}}{j+m} \). But since \(x_{n}\) is the sum of the first \(n\) elements, and \(x_{m}\) is the sum of the first m elements, \(\(x_{n}-x_{m}\)\) is just the sum of the elements from the \((m+1)\) th to the \(n\) th. This verifies that \(x_{n}-x_{m}=(-1)^{m} \sum_{j=1}^{n-m} \frac{(-1)^{j}}{j+m} \).

Separating even and odd terms

Let's separate the even and odd terms of the sum. The even terms have \(j+m\) even and odd terms have \(j+m\) odd. This takes the form \(x_{n}-x_{m}=(-1)^{m}\left\{-\sum_{k=1}^{(n-m) / 2} \frac{1}{2 k-1+m}+\sum_{k=1}^{(n-m) / 2} \frac{1}{2 k+m}\right\}\).

Combining into a single sum

We combine the two sums obtained in Step 2, to form a single sum, leaving us with \(x_{n}-x_{m}=-(-1)^{m}\left\{\sum_{k=1}^{(n-m) / 2} \frac{1}{(2 k+m)(2 k+m-1)}\right\}\), and then analyzing the magnitude of this sum, we get \left|x_{n}-x_{m}\right| \leq \frac{1}{(1+m)^{2}}+\sum_{k=2}^{(n-m) / 2} \frac{1}{(2 k+m-1)^{2}}

Applying integration

In accordance with the given conditions, we can assure that \(\int_{1}^{s} f(x) d x \geq \sum_{k=2}^{s} f(k)\) for any continuous function \(f(x)\). Applying this to part (c), we simplify the expression to get \left|x_{n}-x_{m}\right| \leq \frac{1}{n}+\frac{1}{(1+m)^{2}}-\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{m+1}\right), proving that each term on the last line goes to zero independently as \(m\) and \(n\) go to infinity, hence proving the sequence to be Cauchy.

Key concepts

Series Convergence

In mathematical analysis, understanding whether a series converges is crucial. A series converges if its sequence of partial sums tends towards a specific value as more terms are added. In this exercise, we tackle the sequence defined by the sum of alternately signed reciprocals: \(x_n = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}\). To determine if this sequence is Cauchy, we must show that for any small positive number \(\epsilon\), there exists a positive integer \(N\) such that for all \(n, m > N\), the inequality \(|x_n - x_m| < \epsilon\) holds.
This implies that the terms of the sequence get arbitrarily close to each other as we progress through the sequence. In this case, by decomposing and analyzing the differences \(x_n - x_m\) using series manipulation techniques, we conclude that they tend towards zero as \(n\) and \(m\) grow larger. This signifies that our sequence is indeed Cauchy, hence conveys the series convergence.

Summation Techniques

Summation techniques play a pivotal role in breaking down complex sequences. For the given sequence \(x_n - x_m = (-1)^m \sum_{j=1}^{n-m} \frac{(-1)^j}{j+m}\), analyzing even and odd parts separately becomes handy. First, we identify even and odd indexed elements and then reformulate them into two distinct sequences:

• First series: \(\sum_{k=1}^{(n-m)/2} \frac{1}{2k-1+m}\)
• Second series: \(\sum_{k=1}^{(n-m)/2} \frac{1}{2k+m}\)

By separating and then recombining these parts into a single summation form, the relation \(x_n - x_m = -(-1)^m \sum_{k=1}^{(n-m)/2} \frac{1}{(2k+m)(2k+m-1)}\) is achieved. This kind of manipulation simplifies our calculation and makes it easier to apply convergence assessments, verifying that the terms contribute to a vanishing sum as \(n\) and \(m\) grow large.

Integration in Analysis

In the context of this problem, integration serves as a tool to provide bounds and comparisons with discrete sums. The inequality \(\int_{1}^{s} f(x) \, dx \geq \sum_{k=2}^{s} f(k)\) is pivotal, demonstrating that the integral of a continuous function over a range can approximate or bound the summation over discrete points in that range.
This step is applied to the term \(\left|x_n - x_m\right|\), leading to a bound using integrals. Evaluating \(\int_{1}^{(n-m)/2} \frac{1}{(2x+m-1)^2} \, dx\), yields tighter control over the sum, suggesting each term in this bound approaches zero independently as \(m\) and \(n\) tend toward infinity. Thus, integration techniques solidify our understanding of how discrete summations behave and help establish the fundamental nature of sequence convergence in rigorous mathematical analysis.