Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) $$

Short answer

The given series \(\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)\) diverges.

Step-by-step solution

Simplify the function into a form easy to handle

The given series is \(\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)\).\nWe simplify it by using properties of logarithms. Specifically, we note that \(\ln \left(\frac{n+1}{n}\right)\) is equal to \(\ln(n+1) - \ln(n)\). So, the given series becomes \(\sum_{n=1}^{\infty} (\ln(n+1) - \ln(n))\).

Apply the telescoping series

Telescoping series involves many cancellation of terms when expanded. Expanding it, we get \(\ln2 - \ln1 + \ln3 - \ln2 + \ln4 - \ln3 + ... \). \nWe see that every term cancels another except for -\ln1 in the beginning and \(\ln(n+1)\) in the end, so this is a telescoping series that simplifies to \(\ln(n+1) - \ln1\). When n approaches infinity, \(\ln(n+1)\) goes to ∞. Therefore, the series diverges.

Conclusion

According to the telescoping series property, as n approaches infinity, the series \(\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)\) tends towards infinity. Thus, we conclude that the series diverges.

Key concepts

Convergence and Divergence of Series

Understanding the convergence and divergence of series is essential in calculus, as it determines whether the sum of an infinitely long sequence of numbers settles to a finite value or grows without bound. A series converges if its sequence of partial sums approaches a specific limit. In contrast, a series diverges if it does not converge, meaning that the partial sums do not approach a specific limit.

Several tests can determine convergence or divergence, including the comparison test, the ratio test, and the root test, among others. One particular type of series known as a telescoping series shows a pattern where after expansion, most terms cancel each other out. What remains can give a clear indication of the series' behavior, as was the case in the exercise provided, where after applying properties of logarithms, we observe a telescoping pattern that leads us to deduce the series' divergence.

Properties of Logarithms

Logarithms have properties that make them useful tools in solving mathematical problems, especially when dealing with complex series or sequences. For example, the property \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\) is used to transform a single logarithmic term into a difference of two logarithms. This property is essential when analyzing series like the one in the exercise, where each term \(\ln \left(\frac{n+1}{n}\right)\) is simplified to \(\ln(n+1) - \ln(n)\).

It's vital to understand these properties because they allow us to simplify expressions and reveal underlying patterns, such as the cancellation in telescoping series. Other properties include product rules, power rules, and change of base, all of which provide different ways to manipulate logarithms to our advantage.

Calculus Sequence and Series

In calculus, a sequence is an ordered list of numbers defined by a set rule, and a series is the sum of the terms of a sequence. A crucial part of calculus involves understanding and analyzing sequences and series, especially determining their convergence or divergence. The study of sequences and series not only helps in solving complex mathematical problems but also has practical applications in different fields like physics, computer science, and economics.

When we look at the formula presented in the textbook exercise \(\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)\), we're dealing with an infinite series derived from a sequence. Analyzing such series often involves applying various concepts and techniques from calculus. Recognizing a telescoping series, as done in the step-by-step solution, is just one of the tactics we employ to determine the behavior of complex series.