Problem 2 - Thomas Calculus

Question

In Exercises \(1-6,\) find a formula for the \(n\) th partial sum of each series and use it to find the series' sum if the series converges. $$ \frac{9}{100}+\frac{9}{100^{2}}+\frac{9}{100^{3}}+\cdots+\frac{9}{100^{n}}+\cdots $$

Step-by-Step Solution

Answer

The series converges to \( \frac{1}{11} \).

Step 1: Identify the Series

The series given is an infinite geometric series where the first term \( a = \frac{9}{100} \) and the common ratio \( r = \frac{1}{100} \).

Step 2: Derive the Formula for the n-th Partial Sum

The formula for the \(n\)-th partial sum \(S_n\) of a geometric series is \( S_n = a \frac{1 - r^n}{1 - r} \). Substitute \(a\) and \(r\) into this formula to get \[ S_n = \frac{9}{100} \cdot \frac{1 - \left( \frac{1}{100} \right)^n}{1 - \frac{1}{100}}. \]

Step 3: Simplify the Partial Sum Formula

Simplify the denominator: \( 1 - \frac{1}{100} = \frac{99}{100} \). So, the formula becomes \[ S_n = \frac{9}{100} \times \frac{1 - \left( \frac{1}{100} \right)^n}{\frac{99}{100}} = \frac{9}{100} \times \frac{100}{99} \cdot \left( 1 - \left( \frac{1}{100} \right)^n \right) = \frac{9}{99} \left( 1 - \left( \frac{1}{100} \right)^n \right). \]

Step 4: Determine Convergence of the Series

Since \(-1 < r < 1 \), the series converges. For a geometric series, if \(|r| < 1\), the sum of the series \( S \) is \( \frac{a}{1 - r} \). Substitute the values: \[ S = \frac{\frac{9}{100}}{1 - \frac{1}{100}} = \frac{\frac{9}{100}}{\frac{99}{100}} = \frac{9}{99} = \frac{1}{11}. \]

Key Concepts

Partial Sum Formula

A partial sum in mathematics represents the sum of the first few terms of a series. Specifically, it’s a way to approximate an infinite series by considering only a finite portion of it. For a geometric series — a series where each term is a constant multiple of the previous term — the partial sum formula is especially useful.

The formula for the nth partial sum of a geometric series is expressed as:

• \( S_n = a \frac{1 - r^n}{1 - r} \)

where:

• \( a \) is the first term of the series,
• \( r \) is the common ratio (constant multiple),
• \( n \) is the number of terms being summed.

In the given series, the first term \( a \) is \( \frac{9}{100} \), and the common ratio \( r \) is \( \frac{1}{100} \). By substituting these values into the formula, we find the nth partial sum:

• \( S_n = \frac{9}{100} \cdot \frac{1 - \left( \frac{1}{100} \right)^n}{1 - \frac{1}{100}} \)

This formula simplifies the process of finding a segment of the series, allowing us to evaluate how sums accumulate as more terms are added. Understanding the partial sum formula gives us insight into how infinite series behave as more terms are considered.

Series Convergence

Convergence in the context of a series refers to a condition where adding more terms gradually leads towards a single fixed value. Specifically, for an infinite series to converge, the sequence of its partial sums must approach a specific number as the number of terms increases infinitely.

For geometric series, there is a straightforward criterion to determine convergence:

• If the absolute value of the common ratio \( |r| < 1 \), the series converges.

In our example, \( r = \frac{1}{100} \), which is less than one. This tells us the series will converge because the terms become increasingly smaller toward zero, making the infinite sum stable.

Once confirmed that the series converges, we calculate the sum of the series using the formula:

• \( S = \frac{a}{1 - r} \)

Applying this to our series gives us:

• \( S = \frac{\frac{9}{100}}{1 - \frac{1}{100}} = \frac{1}{11} \)

Thus, convergence not only tells us that a series can be summed but reveals the specific numerical value of that sum.

Infinite Series

An infinite series is a sequence of numbers that continues indefinitely. Each term adds to the sum of all preceding terms, and understanding them is crucial in advanced mathematics and various practical applications.

In the context of this exercise, the infinite series starts as:

• \( \frac{9}{100} + \frac{9}{100^2} + \frac{9}{100^3} + \cdots \)

This series is infinite because it continues without stopping, theoretically accumulating an impossible number of terms. The concept may seem daunting, but it's handled using limits and convergence properties that allow us to make sense of the sum.

To analyze infinite geometric series, we use the rules of convergence and the sum formula discussed earlier. When conditions for convergence are met, even endlessly long series can be tamed into manageable results, like the sum \( \frac{1}{11} \) calculated in this exercise. This transforms what appears infinite into finite, precise insights important in fields from physics to finance.