Question

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height \(n\) is \(1 /(n \ln n)(n \geq 2)\) Assuming that the tower is of infinite height: (a) Will the total area of the disks be finite or not? Hint: Can you compare the series with a simpler one? (b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not? (c) Explain why there is not a contradiction between your answers in (a) and (b). That is, how is it possible to start with a set of disks of finite area, remove a little strip around the circumference of each, and get an infinite total length of these strips? Hint: Think about units-you can't compare area and length. Consider two cases: (1) Make the width of each strip equal to one percent of the radius of the disk from which you cut it. Now the total length is infinite but what about the total area? (2) Try to make the strips all the same width; what happens? Also see Chapter 5, Problem 3.31(b).

Short answer

The total area of the disks is finite, but the total length of the wire required is infinite. There's no contradiction because comparing area (which is finite) and length (infinite) involves different units.

Step-by-step solution

- Formulate the series

The radius at height n is given by \(r_n = \frac{1}{n \ln n}\). The area of each disk is given by the formula for the area of a circle: \(A_n = \pi r_n^2\). Substituting the given radius, the area becomes \(A_n = \pi \left(\frac{1}{n \ln n}\right)^2 = \frac{\pi}{n^2 (\ln n)^2}\). We need to find the sum of the areas for disks at each height, which forms the series \( \sum_{n=2}^{\infty} \frac{\pi}{n^2 (\ln n)^2} \).

- Compare with a known convergent series

To determine if the series \( \sum_{n=2}^{\infty} \frac{\pi}{n^2 (\ln n)^2} \) converges, compare it with the series \( \sum_{n=2}^{\infty} \frac{1}{n^2} \), which is known to converge. Since \( \frac{1}{n^2 (\ln n)^2} < \frac{1}{n^2} \) for all \( n \geq 2 \), and \( \sum_{n=2}^{\infty} \frac{1}{n^2} \) is convergent (a p-series with p = 2, where p > 1), the comparison test indicates that \( \sum_{n=2}^{\infty} \frac{1}{n^2 (\ln n)^2} \) also converges.

- Conclusion for part (a)

Since the series \( \sum_{n=2}^{\infty} \frac{\pi}{n^2 (\ln n)^2} \) converges, the total area of the disks is finite.

- Determine the circumference of each disk

The circumference of each disk is given by \(C_n = 2 \pi r_n = \frac{2 \pi}{n \ln n}\). We need to determine if the series \( \sum_{n=2}^{\infty} \frac{2 \pi}{n \ln n} \) converges or diverges.

- Compare with a known divergent series

To analyze \( \sum_{n=2}^{\infty} \frac{2 \pi}{n \ln n} \), compare it with the harmonic series \( \sum_{n=2}^{\infty} \frac{1}{n} \), which is known to diverge. Since \( \frac{1}{n \ln n} \) is greater than \( \frac{1}{n} \) for all \( n \geq 2 \), and \( \sum_{n=2}^{\infty} \frac{1}{n} \) is divergent, the comparison test indicates that \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \) also diverges.

- Conclusion for part (b)

Since the series \( \sum_{n=2}^{\infty} \frac{2 \pi}{n \ln n} \) diverges, the total length of wire required is infinite.

- Addressing potential contradictions

The apparent contradiction arises from comparing quantities with different units (area vs. length). To see why there's no contradiction, consider two cases: (1) If the width of each strip around the circumference is one percent of the disk's radius, calculate the total area removed. The total area removed is a small percentage of an already convergent series, hence it also converges. Thus, the area remains finite while the total length of the strips is infinite. (2) If each strip has the same physical width, cutting infinitely small strips results in an infinite length. However, the total area removed becomes insignificant compared to the initial area.

Key concepts

Infinite Series

An infinite series is essentially the sum of the terms of an infinite sequence. Understanding whether these sums converge (i.e., approach a finite value) or diverge (i.e., grow indefinitely) is crucial.
In our exercise, we formulated the series representing the total area and total circumference of the disks. Specifically, for the total area, we have:
\( \sum_{n=2}^{\infty} \frac{\pi}{n^2 (\ln n)^2}\fixx \)
For the total length of the wire, we consider the circumference at each height, resulting in the series:
\( \sum_{n=2}^{\infty} \frac{2\pi}{n \ln n}\fix \)
Knowing how to handle and compare these series using convergence tests helps us understand the behavior of these sums as the number of terms grows indefinitely.

Geometry

Geometry is central to understanding this exercise as it involves the shapes and sizes of disks (large coins).
The area of a disk (circle) is given by:
\( A_n = \pi r_n^2 \)
where \(r_n \) is the radius.
Here, the radius for the disks at different heights is given by:
\( r_n = \frac{1}{n \ln n} \).
Similarly, the circumference of a disk, defined as the length around its edge, is:
\( C_n = 2 \pi r_n = \frac{2 \pi}{n \ln n} \).
We use these geometric properties to formulate the series representing the total area and circumference of the disks. This visual and mathematical linkage is powerful for solving the exercise.

Convergence Tests

Convergence tests help us determine if an infinite series converges or diverges. This is pivotal because series with an infinite number of terms can behave differently.
In this exercise, the Comparison Test is utilized.
The Comparison Test involves comparing the given series with a known benchmark series.
For the total area of the disks, we compared: \( \sum_{n=2}^{\infty} \frac{\pi}{n^2(\ln n)^2}\fix \).
We observed \( \frac{1}{n^2 (\ln n)^2}\fix \) is less than \( \frac{1}{n^2}\fix \) which is a known convergent series (p-series with p > 1). Hence, the area series converges.

Comparison Test

The Comparison Test is a powerful method in determining series convergence or divergence.
The idea is to compare a given series with another series whose convergence properties are already known.
In our problem, for the total area of disks: \( \sum_{n=2}^{\infty} \frac{\pi}{n^2(\ln n)^2}\fix \)
We compare it to the p-series: \( \sum_{n=2}^{\infty} \frac{1}{n^2} \).
Since the p-series converges for p=2, it helped establish that our series for area also converges.
For circumference, we compared \( \sum_{n=2}^{\infty} \frac{2\pi}{n \ln n} \) with the harmonic series: \( \sum_{n=2}^{\infty} \frac{1}{n} \), known to diverge.
As \( \frac{2\pi}{n \ln n} \) grows slower than the harmonic series but still diverges, the comparison confirms the total length required is infinite.

Harmonic Series

The Harmonic Series is a fundamental series in mathematics and is defined as: \( \sum_{n=1}^{\infty} \frac{1}{n} \).
Despite its simplicity, it has profound implications.
Notably, this series diverges.
For our problem, the harmonic series helps us understand the behavior of the circumference series.
By comparing: \( \sum_{n=2}^{\infty} \frac{2\pi}{n \ln n} \)
with \( \sum_{n=2}^{\infty} \frac{1}{n} \),
the divergent nature of the harmonic series informs us that the circumference series diverges as well, requiring an infinite total length of wire.

Area versus Length

Understanding the distinction between area and length is key to resolving the apparent contradiction in the problem.
Units play a crucial role.
Area, measured in square units, and length, measured in linear units, cannot be directly compared.
Consider two cases:
1. If the strip around each disk's circumference has a width proportional to the radius (say one percent), the total area removed is proportionally small, hence the area remains finite.
2. If all strips have the same width, although the total length is infinite because of the infinite number of disks, the total area becomes negligible due to the infinitesimally small width.
Thus, explaining how a finite area can coexist with an infinite total length.