Problem 78 - Thomas Calculus

Question

In Exercises \(73-78,\) find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x )\) for those values of \(x .\) $$ \sum_{n=0}^{\infty}(\ln x)^{n} $$

Step-by-Step Solution

Answer

The series converges for \( \frac{1}{e} < x < e \), and the sum is \( S(x) = \frac{1}{1 - \ln x} \).

Step 1: Identify the Geometric Series

The series given is \( \sum_{n=0}^{\infty}(\ln x)^{n} \), which can be recognized as a geometric series with the first term \( a = 1 \) and the common ratio \( r = \ln x \).

Step 2: Determine the Convergence Criteria

A geometric series \( \sum a r^n \) converges if and only if the absolute value of the common ratio \( |r| < 1 \). For our series, this means \( |\ln x| < 1 \).

Step 3: Solve the Inequality

The inequality \( |\ln x| < 1 \) splits into two inequalities: \( \ln x < 1 \) and \( \ln x > -1 \). We solve these inequalities separately.- For \( \ln x < 1 \), exponentiate to get \( x < e^1 = e \).- For \( \ln x > -1 \), exponentiate to get \( x > e^{-1} = \frac{1}{e} \).Therefore, \( x \) must satisfy \( \frac{1}{e} < x < e \).

Step 4: Find the Sum of the Series

If the geometric series converges, the sum is \( S = \frac{a}{1-r} \). Substituting \( a = 1 \) and \( r = \ln x \) gives:\[ S(x) = \frac{1}{1 - \ln x} \]This formula for \( S(x) \) is valid for \( \frac{1}{e} < x < e \).

Key Concepts

Convergence Criteria

In mathematics, particularly in series, understanding when a series converges is crucial. A geometric series is one where each term is a constant multiple of the previous term. For the series to converge, it needs to meet the right criteria. Specifically, the absolute value of the common ratio \( r \) must be less than 1, written as \(|r| < 1\). This condition ensures that the terms of the series keep getting smaller and approach zero, leading to a finite sum.

When dealing with the series \( \sum_{n=0}^{\infty} (\ln x)^n \), the common ratio \( r \) is \( \ln x \). Hence, the series will converge if \(|\ln x| < 1\). This is the foundational step in analyzing whether the series has a finite sum or not.

Sum of Series

Once the convergence criteria are satisfied, we can determine the sum of the series. The sum of an infinite geometric series where the first term is \( a \) and the common ratio is \( r \) (subject to \(|r| < 1\)) is given by the formula:

• \( S = \frac{a}{1 - r} \)

For the geometric series \( \sum_{n=0}^{\infty} (\ln x)^n \), the initial term \( a \) is 1, and the common ratio \( r \) is \( \ln x \). If the series converges, i.e., \( \frac{1}{e} < x < e \), the sum can be calculated using:

• \( S(x) = \frac{1}{1 - \ln x} \)

This formula provides a way to sum the infinite number of terms in the series, as long as the values of \( x \) allow for convergence.

Inequality Solving

It's often necessary to solve inequalities to determine valid values of \( x \) for which a series converges. For the problem \(|\ln x| < 1\), we need to manage two inequalities separately due to the absolute value:

• \( \ln x < 1 \)
• \( \ln x > -1 \)

To solve these, use exponentiation, a method where both sides of the inequality are used as powers of \( e \), the base of natural logarithms.

• For \( \ln x < 1 \), transform it to obtain \( x < e \).
• For \( \ln x > -1 \), result in \( x > \frac{1}{e} \).

Combining these results, the solution to the inequality \(|\ln x| < 1\) is \( \frac{1}{e} < x < e \). This approach narrows down the values of \( x \) to ensure the series converges and the sum formula is applicable.