Question

Is the series convergent or divergent? $$ \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) $$

Short answer

The series diverges as it telescopes to an infinite limit.

Step-by-step solution

Understand the Series Expression

The series given is \( \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) \). This series involves logarithms, so one useful property is \( \ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).

Apply Logarithmic Property

Apply the logarithmic property to each term: \( \ln \left(\frac{n+1}{n}\right) = \ln(n+1) - \ln(n) \). Thus, the series becomes: \( \sum_{n=1}^{\infty} (\ln(n+1) - \ln(n)) \).

Recognize the Telescoping Nature

Notice that the series is telescoping. In a telescoping series, most terms cancel out with subsequent terms. Expanding the first few terms, it becomes clear: \( (\ln(2) - \ln(1)) + (\ln(3) - \ln(2)) + (\ln(4) - \ln(3)) + \ldots \). All intermediate terms cancel.

Evaluate the Remaining Terms

In a telescoping series, after cancellation, the sum simplifies to the first term of the sequence minus its \( n \)-th term as \( n \to \infty \). For this series, it becomes \( \lim_{n \to \infty} (\ln(n+1) - \ln(1)) = \lim_{n \to \infty} \ln(n+1) \).

Determine Convergence by Limit Analysis

As \( n \to \infty \), \( \ln(n+1) \to \infty \). This means the series diverges because the limit of the remaining terms is infinite.

Key concepts

Telescoping Series

A telescoping series is a type of series where many terms cancel each other out, leading to a simplified expression. This cancellation occurs because of the specific structure of the sequence of terms, often involving differences. In our example, the series \( \sum_{n=1}^{\infty} (\ln(n+1) - \ln(n)) \) is telescoping. This is due to the property where each \( \ln(n) \) cancels with the subsequent \( -\ln(n) \) in the successive term.
By expanding the series term-wise, such as \( (\ln(2) - \ln(1)) + (\ln(3) - \ln(2)) + (\ln(4) - \ln(3)) + \ldots \), we observe that most terms vanish, leaving just the start and end terms of the series.
In a telescoping series, what remains after all terms cancel is often very simple and makes it easier to analyze whether a series converges or diverges.

Logarithmic Properties

Logarithmic properties are incredibly useful for manipulating terms in series, making them more manageable. One key property is \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). This property turns division inside a logarithm into subtraction, which simplifies expressions greatly.
In the series \( \sum_{n=1}^{\infty} \ln\left(\frac{n+1}{n}\right) \), applying this property results in \( \sum_{n=1}^{\infty} (\ln(n+1) - \ln(n)) \), clearly showing its telescoping nature.
Recognizing such transformations can reveal structural patterns, like the telescoping pattern, that allow us to determine the convergence properties of a series.

Convergence Tests

Convergence tests are tools used to determine whether an infinite series converges or diverges. For many series, knowing if their sum approaches a finite number or not is crucial.
The series \( \sum_{n=1}^{\infty} \ln\left(\frac{n+1}{n}\right) \) reveals itself as a telescoping series. To determine convergence through cancellation, the series becomes simpler: after cancelling terms, the problem reduces to looking at the limit as \( n \) approaches infinity.
In our problem, evaluating \( \lim_{n \to \infty} \ln(n+1) \) is vital. As \( n \) increases indefinitely, \( \ln(n+1) \) also grows without bound, leading to the conclusion that the series diverges.
Understanding different convergence tests and applying them appropriately, such as the limit test seen here, helps determine the behavior of various series.