Question

Let \(d_{n}\) be an infinite sequence of digits, meaning \(d_{n}\) takes values in \(\\{0,1, \ldots, 9\\}\). What is the largest possible value of \(x=\sum_{n=1}^{\infty} d_{n} / 10^{n}\) that converges?

Short answer

The largest possible value of the sequence is 1.

Step-by-step solution

Understanding the Series

The series is given by the formula \(x = \sum_{n=1}^{\infty} \frac{d_{n}}{10^{n}}\), where each \(d_n\) is a digit ranging from 0 to 9. This series is similar to a decimal representation.

Maximizing Each Term

To maximize the value of \(x\), we need to maximize each \(d_n\), as each term \(\frac{d_{n}}{10^{n}}\) contributes to \(x\). The maximum value for each \(d_n\) is 9.

Constructing the Maximum Series

If each \(d_n\) is set to 9, the series becomes \(\sum_{n=1}^{\infty} \frac{9}{10^{n}}\). This can be rewritten as \(9 \sum_{n=1}^{\infty} \frac{1}{10^{n}}\).

Calculating the Sum of a Geometric Series

The series \(\sum_{n=1}^{\infty} \frac{1}{10^{n}}\) is a geometric series with the first term \(a = \frac{1}{10}\) and common ratio \(r = \frac{1}{10}\). The sum of an infinite geometric series is given by \(\frac{a}{1-r}\).

Applying the Formula for Infinite Series

Using the formula \(\frac{a}{1-r}\), we calculate the sum: \(\frac{\frac{1}{10}}{1-\frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9}\).

Finding the Complete Solution

Thus, the maximum value of \(x\) is \(9 \times \frac{1}{9} = 1\).

Key concepts

Geometric Series

In mathematics, a geometric series is a series of terms where each term after the first is found by multiplying the previous one by a constant called the common ratio. For the series to be a geometric series, this pattern must hold throughout the sequence.
A geometric series is represented as:

• The first term, denoted by \(a\), decides the starting point.
• The common ratio, \(r\), less than 1 ensures each term gets progressively smaller in the context of convergence.

For a geometric series \[a, ar, ar^2, ar^3, \ldots \] the sum of an infinite geometric series can be written as: \[S = \frac{a}{1-r}\] provided that \(|r| < 1\). This formula provides the convenience of summing up the series without calculating every term. For example, in our series,

• \(a = \frac{1}{10}\)
• \( r = \frac{1}{10}\)
• The sum would be \(\frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{1}{9}\)

Decimal Representation

Decimal representation is a way of naturally expressing numbers using a base 10 system. Each digit in a decimal number is an integer value between 0 and 9. Numbers are often expressed as the sum of terms of decreasing powers of 10.
For any given number \(x\), represented by \(x = d_0 + \frac{d_1}{10} + \frac{d_2}{100} + \frac{d_3}{1000} + \ldots\), where each \(d_n\) is a decimal digit, the representation extends potentially indefinitely in the context of decimal numbers between 0 and 1. In this setup, for the maximum digit of each place value, you choose the digit 9. So for the sequence we explore:

• \(x = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots\)

This reflects a pattern reminiscent of decimals like 0.9999..., which actually mathematically equates to 1.

Convergence of Series

Convergence in the context of series implies that the sum of the series approaches a specific number as more terms are added. Specifically, an infinite series converges if the sequence of its partial sums approaches a finite limit. Convergence ensures that even though we're adding an infinite number of terms, they effectively "add up" to an identifiable sum. For geometric series, convergence happens when the absolute value of the common ratio is less than 1. This means each successive term gets smaller, contributing less to the overall sum, and thus allowing it to settle towards a fixed value.
For instance, the series \(\sum_{n=1}^{\infty} \frac{1}{10^n}\) converges because:

• The common ratio \( r = \frac{1}{10} \) ensures the terms decrease in magnitude.
• As we add more terms, they become negligible in size, leading the sum towards a specific limit.

In our case, we've seen that the series converges to \(\frac{1}{9}\). When multiplied by 9, as seen in the original problem, it reaches the number 1, reflecting a concept that might initially seem counterintuitive.