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Chapter 3: Problem 10

Show that $$ \sum_{i \in \mathbb{N}} \frac{1}{i} $$ diverges. Are there any infinite subsets \(J \subset \mathbb{N}\) such that $$ \sum_{i \in J} \frac{1}{i} $$ converges?

Short Answer

Expert verified
The harmonic series diverges. Infinite subsets like perfect squares can make the series converge.

Step by step solution

01

Recall the Harmonic Series

The series \( \sum_{i=1}^{\infty} \frac{1}{i} \) is known as the harmonic series. An important property of this series is that it diverges. This means that as we add more and more terms, the sum increases without bound.
02

Apply Divergence Test

To show that the harmonic series diverges, we can use the divergence test. This test states that if the limit of the terms \( a_i \) is non-zero as \( i \to \infty \), then the series \( \sum a_i \) must diverge. For the harmonic series, the terms are \( \frac{1}{i} \), and \( \lim_{i \to \infty} \frac{1}{i} = 0 \). However, the harmonic series is known to diverge, prompting us to use an additional argument.
03

Compare with Integral

To further illustrate divergence, we can compare the harmonic series to the integral \( \int_{1}^{N} \frac{1}{x} \, dx \). This integral evaluates to \( \ln(N) \), which clearly tends to infinity as \( N \to \infty \). Since the integral diverges, the harmonic series \( \sum_{i=1}^{\infty} \frac{1}{i} \) also diverges.
04

Identify Subsets for Convergence

There are indeed infinite subsets \( J \subset \mathbb{N} \) such that \( \sum_{i \in J} \frac{1}{i} \) converges. An example is the set of perfect squares \( J = \{i^2 \mid i \in \mathbb{N} \} \). Since \( \sum_{i=1}^{\infty} \frac{1}{i^2} \) converges (known from the p-series test with \( p = 2 \)), selecting terms from such subsets can yield a convergent series.

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Key Concepts

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

Divergence Test
When evaluating whether a series converges or diverges, the divergence test is an essential tool. The divergence test can quickly show that a series must diverge. The key element of this test is simple: if the limit of the terms of a series does not equal zero, then the series diverges. This helps identify series where further analysis will be necessary if our direct application yields zero.
Let's focus on its syntax: consider a series \( \sum a_i \). Using the divergence test, if \( \lim_{i \to \infty} a_i eq 0 \), then \( \sum a_i \) diverges. However, always note that if the limit is zero, this does not guarantee convergence. Take the harmonic series, for example, which has terms \( \frac{1}{i} \) that tend to zero, but the series itself diverges.
Integral Comparison
The integral comparison method is beneficial for evaluating the behavior of series, especially those like the harmonic series. At the heart of this method is comparing the series to an integral, which can easier reveal divergence or convergence.
To apply this, consider the harmonic series \( \sum_{i=1}^{\infty} \frac{1}{i} \). We can compare this to the integral \( \int_{1}^{N} \frac{1}{x} \, dx \). Calculating this integral results in \( \ln(N) \), which approaches infinity as \( N \to \infty \). Therefore, because the integral diverges, so does the series. This technique is particularly effective when dealing with series that might not have straightforward algebraic solutions.
P-Series Test
The p-series test is a vital concept for series analysis. It helps determine whether a series converges or diverges, based on its exponent's value. A p-series takes the form \( \sum_{i=1}^{\infty} \frac{1}{i^p} \). The outcome depends on \( p \), the exponent.
The rule is straightforward: the series converges if \( p > 1 \), and diverges if \( p \leq 1 \). This reveals why the harmonic series \( \sum_{i=1}^{\infty} \frac{1}{i} \), a p-series with \( p = 1 \), diverges. Conversely, a series like \( \sum_{i=1}^{\infty} \frac{1}{i^2} \) converges because \( p = 2 \). This test is instrumental for analysing subsets and more complex series analysis.

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Most popular questions from this chapter

Let \(\left\\{a_{n}\right\\}\) be a sequence of real numbers converging to zero. Show that there must exist a monotonic sequence \(\left\\{b_{n}\right\\}\) such that the series \(\sum_{k=1}^{\infty} b_{k}\) diverges and the series \(\sum_{k=1}^{\infty} a_{k} b_{k}\) is absolutely convergent.

A sequence \(\left\\{x_{n}\right\\}\) of real numbers is said to be of bounded variation if the series $$ \sum_{k=2}^{\infty}\left|x_{k}-x_{k-1}\right| $$ converges. (a) Show that every sequence of bounded variation is convergent. (b) Show that not every convergent sequence is of bounded variation. (c) Show that all monotonic convergent sequences are of bounded variation. (d) Show that any linear combination of two sequences of bounded variation is of bounded variation. (e) Is the product of of two sequences of bounded variation also of bounded variation?

This collection of exercises develops some convergence properties of trigonometric series; that is, series of the form $$ a_{0} / 2+\sum_{k=1}^{\infty}\left(a_{k} \cos k x+b_{k} \sin k x\right) $$ (a) For what values of \(x\) does \(\sum_{k=1}^{\infty} \frac{\sin k x}{k^{2}}\) converge? (b) For what values of \(x\) does \(\sum_{k=1}^{\infty} \frac{\sin k x}{k}\) converge? (c) Show that the condition \(\sum_{k=1}^{\infty}\left(\left|a_{k}\right|+\left|b_{k}\right|\right)<\infty\) ensures the absolute convergence of the trigonometric series (10) for all values of \(x\).

If \(\sum_{k=1}^{\infty}\left(a_{k}+b_{k}\right)\) diverges, what can you say about the series $$ \sum_{k=1}^{\infty} a_{k} \text { and } \sum_{k=1}^{\infty} b_{k} ? $$

Suppose that \(\sum_{n=1}^{\infty} a_{n}\) is a convergent series of positive terms. Must the series \(\sum_{n=1}^{\infty} \sqrt{a_{n}}\) also be convergent?
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