Problem 32 - Calculus: One and Several Variables

Question

(a) Prove that if \(\sum_{k=0}^{\infty} a_{k}\) is a convergent series with all terms nonzero, then \(\sum_{x=0}^{\infty}\left(1 / a_{k}\right)\) diverges. (b) Suppose that \(a_{k}>0\) for all \(k\) and \(\sum_{k=1}^{\infty} a_{k}\) diverges. Show by example that \(\sum_{k=0}^{\infty}\left(1 / a_{k}\right)\) may converge and it may diverge.

Step-by-Step Solution

Answer

In summary, for part (a) we proved that if a series \(\sum_{k=0}^{\infty} a_{k}\) converges with all terms nonzero, then its reciprocal series \(\sum_{k=0}^{\infty} \left(\frac{1}{a_{k}}\right)\) diverges. For part (b), we found two examples: the Harmonic series, which diverges while its reciprocal series converges, and the series \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\), which diverges along with its reciprocal series.

Step 1: Part (a): Prove that the reciprocal series diverges

Let's consider a convergent series \(\sum_{k=0}^{\infty} a_{k}\) with all terms nonzero. Since it is convergent, we can say that its limit is finite: \[\lim_{n \to \infty} \sum_{k=0}^{n} a_{k} = L < \infty\] Now we will look at the reciprocal series, and assume it converges: \[\lim_{n \to \infty} \sum_{k=0}^{n} \left(\frac{1}{a_{k}}\right) = M\] We will try to show that this assumption leads to a contradiction. To do this, let's take the limit of their product: \[\lim_{n \to \infty} \left(\sum_{k=0}^{n} a_{k}\right) \left(\sum_{k=0}^{n} \frac{1}{a_{k}}\right)\] Since the limits of both series exist, we can say: \[\lim_{n \to \infty} \left(\sum_{k=0}^{n} a_{k}\right) \left(\sum_{k=0}^{n} \frac{1}{a_{k}}\right) = L \cdot M\] However, if we expand the product, we obtain the sum of reciprocal pairs: \[\sum_{k=0}^{n} \left(a_{k} \cdot \frac{1}{a_{k}}\right) = \sum_{k=0}^{n} 1 = n + 1\] Taking the limit: \[\lim_{n \to \infty} (n + 1) = \infty\] This is a contradiction because we assumed that the limit of the product is finite (L * M). Therefore, our initial assumption that the reciprocal series converges must be false, and the reciprocal series is, in fact, divergent.

Step 2: Part (b): Finding examples of the series

We need to find an example of a series that diverges, but its reciprocal series converges, and another example where both diverge. 1. Example with divergent series and convergent reciprocal series: Let's take the Harmonic series: \[\sum_{k=1}^{\infty} \frac{1}{k}\] It is well-known that the Harmonic series diverges. The reciprocal series will be: \[\sum_{k=1}^{\infty} k\] This series converges to a single point since it is a geometric series with a constant ratio less than 1. Hence, the series' limit is finite, and it converges. 2. Example with both divergent series and divergent reciprocal series: Let's consider the series: \[\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\] This series diverges due to the p-series test (p = 1/2 < 1). The reciprocal series will be: \[\sum_{k=1}^{\infty} \sqrt{k}\] This series diverges as well because the terms do not tend to 0 (each term grows without bound). Therefore, the reciprocal series is also divergent in this case. So, we have shown two examples for part (b), one with a divergent series and convergent reciprocal series, and another where both diverge.

Key Concepts

Reciprocal Series

A reciprocal series arises when we take the terms of an existing series and replace each term with its reciprocal. Consider a series \( \sum_{k=0}^{\infty} a_{k} \). If all of its terms, \( a_{k} \), are nonzero, the corresponding reciprocal series is represented as \( \sum_{k=0}^{\infty} \frac{1}{a_{k}} \). This transforms each term from \( a_{k} \) to \( \frac{1}{a_{k}} \).

In step 1 of our solution, we proved that for a convergent series \( \sum_{k=0}^{\infty} a_{k} \) with nonzero terms, the reciprocal series \( \sum_{k=0}^{\infty} \frac{1}{a_{k}} \) must diverge. This happens because when attempting to multiply the original convergent series by its reciprocal series, the result trends towards infinity, rather than staying finite. This results in a contradiction, indicating divergence.

Understanding reciprocal series can be vital when analyzing the behavior of original series in mathematical or applied contexts.

Divergent Series

A divergent series is one where the sum of its terms does not approach a finite limit. In other words, as you add more terms, the total either increases indefinitely or doesn't settle down to a particular value. The harmonic series, for example, \( \sum_{k=1}^{\infty} \frac{1}{k} \), is a classic divergent series. Even though each term becomes smaller as \( k \) increases, the series itself grows without bound in the long run.

In the given solution, Part (b) explores divergent series with a concretized perspective, where examples highlight different behaviors. Sometimes, a divergent series can have a reciprocal series that converges. For example, the divergent harmonic series' reciprocal, which is the geometric series \( \sum_{k=1}^{\infty} k \), interestingly converges.

On the other hand, the series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \) and its reciprocal, \( \sum_{k=1}^{\infty} \sqrt{k} \), both diverge. These illustrations emphasize that knowing a series diverges doesn't always clearly predict the behavior of its reciprocal.

Limit Comparison Test

The limit comparison test is a powerful tool used in mathematics to determine the convergence or divergence of a series by comparing it with another series whose behavior is already known. The test involves examining the limit of the ratio of the terms of two series. If we have two series, \( \sum a_k \) and \( \sum b_k \), and we want to know if \( \sum a_k \) converges or diverges, we compute \[ \lim_{k \to \infty} \frac{a_k}{b_k} = c \]

If the limit \( c \) is finite and positive, both series either converge or diverge together.

In the context of the given problem, although the direct use of limit comparison was not detailed, understanding its application can give students more insight into how certain series behave. For instance, when analyzing divergent series and their reciprocals, the intuition from limit comparison helps predict behavior based on known convergent or divergent series.

As such, having the foundational knowledge of the limit comparison test allows learners to tackle broad scenarios by relating unfamiliar series to familiar ones using established limits.