Problem 45 - Thomas Calculus

Question

Is it true that if \(\sum_{n=1}^{\infty} a_{n}\) is a divergent series of positive numbers, then there is also a divergent series \(\sum_{n=1}^{\infty} b_{n}\) of positive numbers with \(b_{n}<\) \(a_{n}\) for every \(n ?\) Is there a smallest divergent series of positive numbers? Give reasons for your answers.

Step-by-Step Solution

Answer

No, not every smaller positive series diverges; there's no smallest divergent series of positive numbers.

Step 1: Understanding Divergent Series

A series \( \sum_{n=1}^{\infty} a_n \) is divergent if the partial sums \( S_n = \sum_{k=1}^{n} a_k \) do not converge to a finite limit. Since \( a_n \) are positive, this means the partial sums grow without bound as \( n \) increases.

Step 2: Using Comparison Test for Series

If \( \sum_{n=1}^{\infty} a_n \) diverges and \( 0 < b_n < a_n \), we cannot automatically conclude that \( \sum_{n=1}^{\infty} b_n \) must also diverge. Because the series of \( b_n \) is smaller term-by-term, it might converge even if \( a_n \) diverges. The Comparison Test does not tell us about divergence of smaller series.

Step 3: Existence of Smaller Divergent Series

Consider the series \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is a known divergent series (harmonic series). Any series \( b_n \) that satisfies \( b_n = \frac{1}{n + 1} \) will have \( b_n < a_n = \frac{1}{n} \) for all \( n \geq 1 \), yet remains divergent. Adding a constant factor (less than 1) such as \( b_n = \frac{1}{2n} \) can make the series converge.

Step 4: Discussion of Smallest Divergent Series

There is no 'smallest' divergent series. Divergence is defined by the behavior of the partial sums (growing unbounded). Changing a constant factor in a divergent sequence \( a_n \) cannot create a new 'image', only changing the nature of sum's divergence. A 'smallest' series doesn't seem meaningful since divergences are usually not bound except contextual limitations.

Key Concepts

Comparison Test

The Comparison Test is a useful tool for determining the convergence or divergence of series. Its basic premise is to compare a given series to another series whose convergence properties are already known. If you have two series, \[\sum_{n=1}^{\infty} a_n\] and \[\sum_{n=1}^{\infty} b_n\], the Comparison Test allows you to make conclusions about their behavior by comparing their terms, \(a_n\) and \(b_n\), for each term in the series.

Here's how it works:

• If \( 0 \leq b_n \leq a_n \) for all \(n\), and \( \sum_{n=1}^{\infty} a_n\) converges, then \( \sum_{n=1}^{\infty} b_n \) also converges.
• Conversely, if \(0 \leq a_n \leq b_n\) for all \(n\), and \(\sum_{n=1}^{\infty} b_n\) diverges, then \(\sum_{n=1}^{\infty} a_n\) also diverges.

However, the Comparison Test doesn't provide information about divergence when the smaller series diverges but isn’t large enough relative to the original series. For example, if \(\sum_{n=1}^{\infty} a_n\) diverges and \(0 < b_n < a_n\), it doesn't guarantee that \(\sum_{n=1}^{\infty} b_n\) diverges.
In situations with positive terms where \(b_n < a_n\), a additional tests are needed to conclude whether \(\sum_{n=1}^{\infty} b_n\) diverges or converges.

Harmonic Series

The harmonic series, \[\sum_{n=1}^{\infty} \frac{1}{n},\] is a classic example of a divergent series. Each term in the harmonic series is the reciprocal of an integer. Although these terms get smaller as \(n\) increases, their sum grows without bound.

This is an important concept because it shows how a series with individual terms tending towards zero can still diverge. The divergence of the harmonic series demonstrates that there is no necessary "smallest" divergent series, as even seemingly decreasing series like the harmonic one continue to grow indefinitely.

The harmonic series serves as a benchmark in the study of divergence and comparison. This series can be manipulated to explore the impact of changes to each term's value. As an example:

• Consider \(b_n = \frac{1}{n+1}\): while this differs from the harmonic sequence, it remains divergent.
• If \(b_n = \frac{1}{2n}\), the series could converge since the terms rapidly decrease.

This series helps to clarify why some divergent series do not necessarily have an apparent smallest value.

Partial Sums

Partial sums are the sums of the first \(n\) terms of a series, represented by \( S_n = \sum_{k=1}^{n} a_k \). They are essential in determining the convergence or divergence of a series.

If the partial sums \( S_n \) tend towards a finite number as \(n \rightarrow \infty\), the series converges. However, if \( S_n \) grows without bound, the series diverges.

Understanding partial sums can help explain why there isn't a smallest divergent series. Consider the behavior of partial sums in divergent series like the harmonic series:

• The partial sums of the harmonic series, \(S_n = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}\), increase without limit as more terms are added.
• Adjusting terms like in the series \(\sum_{n=1}^{\infty} \frac{1}{n+1}\) also results in the partial sums growing infinitely, keeping the series divergent.

Thus, when evaluating series with positive terms, attention to how their partial sums behave gives insight into their divergence.