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. $$\frac{9}{100}+\frac{9}{100^{2}}+\frac{9}{100^{3}}+\dots+\frac{9}{100^{n}}+\cdots$$

Step-by-Step Solution

Answer

The series converges, and its sum is \( \frac{1}{11} \).

Step 1: Identify the Type of Series

The series given is \( \frac{9}{100} + \frac{9}{100^2} + \frac{9}{100^3} + \cdots \). This is a geometric series because each term is obtained by multiplying the previous term by a constant ratio. We can identify the first term \( a = \frac{9}{100} \) and the common ratio \( r = \frac{1}{100} \).

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

For a geometric series, the formula for the n-th partial sum \( S_n \) is given by:\[S_n = a \frac{1-r^n}{1-r}\]where \( a \) is the first term and \( r \) is the common ratio.

Step 3: Substitute Values into the Formula

Substitute \( a = \frac{9}{100} \) and \( r = \frac{1}{100} \) into the partial sum formula:\[S_n = \frac{9}{100} \frac{1-(\frac{1}{100})^n}{1-\frac{1}{100}}\]

Step 4: Simplify the Formula for the n-th Partial Sum

Simplify the denominator:\[1 - \frac{1}{100} = \frac{99}{100}\]Thus, the formula becomes:\[S_n = \frac{9}{100} \cdot \frac{1-(\frac{1}{100})^n}{\frac{99}{100}} = \frac{9}{99} (1-(\frac{1}{100})^n)\]Further simplifying gives us:\[S_n = \frac{1}{11} (1-(\frac{1}{100})^n)\]

Step 5: Determine Convergence and Find the Sum

Since the absolute value of the common ratio \( r=\frac{1}{100} \) is less than 1, the series converges. The sum of the infinite series \( S \) is given by the formula for an infinite geometric series:\[S = \frac{a}{1-r} = \frac{\frac{9}{100}}{1-\frac{1}{100}} = \frac{9}{99} = \frac{1}{11}\]

Key Concepts

Partial Sum

A partial sum in a series means adding up a specific number of terms from the beginning up to a certain term, which we call the n-th term. In the context of a geometric series, a partial sum lets us observe how the series accumulates over its first few terms. By looking at partial sums, we can gain an idea of what might happen as we add more terms into infinity.
For our series, the partial sum formula is:

• First, find the first term of the series, denoted as \(a\). For our series, \(a = \frac{9}{100}\).
• Next, identify the common ratio \(r\), which is \(\frac{1}{100} \). This ratio tells us how each term relates to the one before it.
• The formula for the n-th partial sum \( S_n \) of a geometric series is given by:\[S_n = a \frac{1-r^n}{1-r}\]

To find \(S_n\) for many different values of \(n\), we substitute \(a\) and \(r\) into this formula. This helps us see how the sum grows as \(n\) increases.

Series Convergence

Series convergence is a concept involving whether adding infinitely many terms of a series will approach a finite limit or not. For a series to converge, the terms added must get closer and closer to zero as we progress. In simpler terms, for a geometric series, we are checking whether the total sum settles on a particular number when we add an infinite number of terms.
A geometric series with a common ratio \(r\) converges if

• The absolute value of \(r\) is less than 1, i.e., \(|r| < 1\).

For our series:

• The common ratio \(r = \frac{1}{100}\) which satisfies our convergence condition, since \(| \frac{1}{100} | = 0.01 < 1\).
• Thus, we can expect the series sum to approach a finite number as we add more terms.

Under this condition, the infinite series has an exact sum.

Infinite Series

An infinite series sums infinitely many terms. In a geometric series, if the series converges, it sums to an exact finite number despite having infinite terms. This phenomenon can seem counterintuitive because we are dealing with a never-ending process.
The sum of an infinite geometric series can be calculated with the formula:

• To find the sum \(S\), use:\[S = \frac{a}{1-r}\]Where \(a\) is the first term, and \(r\) is the common ratio.
• Our series simplifies to \(S = \frac{9/100}{1-1/100} = \frac{9}{99} = \frac{1}{11}\).

Though terms keep adding indefinitely, they collectively settle down to the number \(\frac{1}{11}\) indicating a neat total, achievable because each added term shrinks by the constant factor \(r\).

Common Ratio

The common ratio in a geometric series is the constant we multiply by to transition from one term to the next. It plays a crucial role in defining the entire series and its behavior.
Detecting the common ratio:

• In our given series, each subsequent term is derived by multiplying the previous term by \(\frac{1}{100} \), the common ratio \(r\).
• This common ratio determines if a geometric series converges or diverges.
• The rule for convergence is simple: if the absolute value of the common ratio \(|r|\) is less than one, the series converges, meaning it adds up to a fixed sum eventually.

For this series, since \(r=\frac{1}{100}\), the series converges gracefully towards a finite sum. This consistency in multiplying by a constant factor illustrates the predictable pattern characteristic of geometric series.