Question

In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from \(\sum_{n=1}^{\infty} \frac{1}{n}\) by removing any term \(1 / n\) if a given digit, say 9, appears in the decimal expansion of \(n\). Argue that this depleted harmonic series converges by answering the following questions. a. How many whole numbers \(n\) have \(d\) digits? b. How many \(d\) -digit whole numbers \(h(d)\). do not contain 9 as one or more of their digits? c. What is the smallest \(d\) -digit number \(m(d) ?\) d. Explain why the deleted harmonic series is bounded by \(\sum_{d=1}^{\infty} \frac{h(d)}{m(d)}\) e. Show that \(\sum_{d=1}^{\infty} \frac{h(d)}{m(d)}\) converges.

Short answer

The depleted harmonic series converges because its bounding series is geometric with ratio \( r = \frac{9}{10} \).

Step-by-step solution

Calculate Total d-digit Numbers

To find the number of whole numbers with \( d \) digits, we calculate the difference between the smallest \( d \)-digit number and the smallest \( (d-1) \)-digit number. The smallest \( d \)-digit number is \( 10^{d-1} \), and the largest is \( 10^d - 1 \). Thus, there are \( 10^d - 10^{d-1} = 9 \times 10^{d-1} \) whole numbers with \( d \) digits.

Determine the Number of d-digit Numbers Without 9

If a \( d \)-digit number cannot contain the digit 9, it can only use the other digits (0-8). The first digit of a \( d \)-digit number cannot be zero, so it has 9 possible choices (1-9), and each of the remaining \( d-1 \) digits has 9 options (0-8). Hence, the number of \( d \)-digit numbers without a 9 is \( 9 \times 9^{d-1} = 9^d \). This answers the question of how many \( d \)-digit whole numbers \( h(d) \) do not contain 9.

Identify the Smallest d-digit Number

The smallest \( d \)-digit number \( m(d) \) is \( 10^{d-1} \). For example, when \( d = 1 \), \( m(d) = 1 \); when \( d = 2 \), \( m(d) = 10 \), and so on.

Explain Why the Deleted Series is Bounded

The deleted harmonic series excludes terms where the indices have the digit 9, reducing the series to terms where the number does not include 9. Hence, the series can be expressed as a sum involving only those numbers, represented as \( \sum_{d=1}^{\infty} \frac{h(d)}{m(d)} \), where \( h(d) \) are terms without 9 and \( m(d) \) appropriately scales by digit length. This forms a well-defined upper bound.

Show Convergence of Bounding Series

To demonstrate convergence of \( \sum_{d=1}^{\infty} \frac{h(d)}{m(d)} = \sum_{d=1}^{\infty} \frac{9^d}{10^{d-1}} \), notice that this is equivalent to \( \sum_{d=1}^{\infty} 9 \left(\frac{9}{10}\right)^d \), which is a geometric series. A geometric series \( \sum ar^n \) converges if \(|r| < 1\). Here, \( r = \frac{9}{10} \) and thus \(|r| < 1\), ensuring convergence.

Key concepts

Harmonic Series

The harmonic series is a crucial concept in mathematics that refers to the infinite series \( \sum_{n=1}^{\infty} \frac{1}{n} \). In simple terms, it is the sum of the reciprocals of all positive integers. This series is notably famous because, despite the terms getting smaller and smaller as \( n \) increases, the series itself does not converge. It keeps adding up infinitely, albeit very slowly.
The divergence of the harmonic series is a fundamental result, making it an important example when discussing series convergence or divergence. It's helpful to remember that even though each individual term is getting closer to zero, the endless accumulation leads to infinity. This shows how not all series with terms approaching zero will converge.
Understanding why the harmonic series diverges can help you when learning about other more complex series.

Depleted Series

A depleted series is derived by removing certain terms from a base series. In our specific context, the depleted harmonic series is formed by excluding terms that meet a specific criterion from the harmonic series. For example, you can remove all terms where the number contains the digit 9. When terms are removed in this manner, it can significantly alter the convergence properties of the series.
In the instance of the depleted harmonic series discussed, the act of eliminating terms where the number includes a 9 results in a series that does converge. This happens because effectively, you truncate the harmonic series to a point where the combined sum of the remaining terms forms a convergent series.
This illustrates an interesting property: by altering the sequence in a precise way, such as removing a pattern of terms, you can change a divergent series into a convergent one.

Geometric Series

The geometric series is a simple yet profound type of series in mathematics given by \( \sum ar^n \). It consists of terms that are powers of a fixed ratio \( r \). This series is pivotal because it provides one of the clearest examples of a convergent series.
For a geometric series to converge, the absolute value of the common ratio \( r \) must be less than 1. This is because if \( |r| < 1 \), the terms shrink exponentially, leading to a series that sums to a finite value.
In the exercise solution, the series formed by the depleted harmonic series aligns with a geometric series where \( r = \frac{9}{10} \). Since \( |r| < 1 \), this series converges. This example highlights the importance of identifying the type of series you are working with, as it directly influences the behavior and convergence of the series.

Bounded Series

When dealing with series like the depleted harmonic series, understanding the concept of bounds is pivotal. A bounded series is one where the sum of terms does not exceed a certain upper limit, despite adding potentially infinite terms.
In the solution context, the depleted harmonic series is described as being bounded by another series \( \sum_{d=1}^{\infty} \frac{h(d)}{m(d)} \). This means the total sum of the depleted series does not exceed the calculated sum of this bounding series.
Knowing a series is bounded provides assurance about the extent of its values. In our case, this bounded nature is partially why the series converges, because not only are the terms becoming smaller, but they're also capped by a clearly defined upper boundary, limiting their cumulative growth.