91Ó°ÊÓ

To determine whether a geometric series converges, we need to examine the convergence criteria for such series. A geometric series has the form \( a + ar + ar^2 + \cdots \), where \( a \) is the initial term and \( r \) is the common ratio. The key condition for convergence is based on the value of \( r \). The series converges if and only if the absolute value of the common ratio \( |r| < 1 \). This means that the terms in the series get progressively smaller and approach zero, ensuring the sum of the series settles at a certain value rather than growing indefinitely.
In our exercise, the geometric series is expressed as \( \sum_{n=0}^{\infty} (\ln x)^n \), with the first term \( a = 1 \) and common ratio \( r = \ln x \). For this series to converge, we must have \( |\ln x| < 1 \). This inequality ensures that the series terms decrease to zero, allowing the sum to converge to a finite value.

Once a geometric series is determined to converge, we can calculate the sum of the series by using the formula for the sum of an infinite geometric series. The sum is given by \( \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. This formula stems from the fact that as the terms diminish to zero, the series effectively becomes a manageable expression.
For our series where \( a = 1 \) and \( r = \ln x \), the sum becomes \( \frac{1}{1-\ln x} \). However, it's crucial to remember that this sum is only valid for the range of \( x \) that satisfies the convergence criteria, specifically \( \frac{1}{e} < x < e \). In this interval, the logarithmic term stays within the bounds necessary for convergence, and the sum \( \frac{1}{1-\ln x} \) remains a real number. If \( x \) falls outside of this range, the series does not converge, and the sum formula is not applicable.

To solve the convergence criterion \( |\ln x| < 1 \), we translate this into two inequalities: \( \ln x < 1 \) and \( \ln x > -1 \). Working through these inequalities helps us find the suitable range of \( x \) values for convergence.
First, solving \( \ln x < 1 \) requires exponentiating both sides to derive \( x < e \), since the exponential function is the inverse of the natural logarithm. Next, solving \( \ln x > -1 \) by exponentiating both sides gives us \( x > \frac{1}{e} \). When combined, these inequalities define the range \( \frac{1}{e} < x < e \).

• This range is essential as it restricts \( x \) to values where the log expression remains within the boundaries of \(-1\) and \(1\), ensuring that the series converges.

This method of exponential transformation simplifies solving logarithmic inequalities, a vital tool in calculus for understanding the behavior of functions and series.