Question

Express the following series as a telescoping sum and evaluate its nth partial sum. $$ \sum_{n=1}^{\infty} \ln \left(\frac{n}{n+1}\right) $$

Short answer

The series diverges to \(-\infty\) as \( n \to \infty \).

Step-by-step solution

Identify the Structure of the Series

The given series is \( \sum_{n=1}^{\infty} \ln \left(\frac{n}{n+1}\right) \). We observe that the argument of the logarithm can be rewritten using logarithmic properties as \( \ln(n) - \ln(n+1) \). This split is the defining form of a telescoping series.

Write the Series in Telescoping Form

Represent the series as a sum of differences: \( \sum_{n=1}^{\infty} \left( \ln(n) - \ln(n+1) \right) \). This form makes it easier to identify cancellation of terms when summed.

Write the Partial Sum and Observe Cancellation

The nth partial sum \( S_n \) is given by \( S_n = \sum_{k=1}^{n} \left( \ln(k) - \ln(k+1) \right) \). Write out the first few terms: \( (\ln(1) - \ln(2)) + (\ln(2) - \ln(3)) + \ldots + (\ln(n) - \ln(n+1)) \). Notice how terms cancel: after cancellation, \( S_n = \ln(1) - \ln(n+1) \).

Simplify the Expression for the nth Partial Sum

Simplify \( S_n = \ln(1) - \ln(n+1) = 0 - \ln(n+1) = -\ln(n+1) \). Thus, the nth partial sum is \( -\ln(n+1) \).

Evaluate the Infinite Sum

To find the sum of the infinite series, consider the limit of the partial sums as \( n \to \infty \): \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} -\ln(n+1) \). Since \( \ln(n+1) \to \infty \), \( -\ln(n+1) \to -\infty \). Hence, the series diverges to \(-\infty\).

Key concepts

Understanding Partial Sums

When working with series, especially those involving infinite terms, often, we first look at partial sums to see how the series behaves. The "partial sum" is simply the sum of the first few numbers in the series. For the series given in the exercise, the partial sum is represented as:

• For the nth partial sum, we add up terms: \( S_n = \ln(1) - \ln(2) + \ln(2) - \ln(3) + \ldots + \ln(n) - \ln(n+1) \).
• Key thing to notice: When terms are written out, many inner terms cancel each other out.
• Only the first term from the first part and the last term from the second part remain in the summation.

This leads to a simplified expression for the nth partial sum as \( S_n = \ln(1) - \ln(n+1) \). Here, \( \ln(1) = 0 \), which further simplifies to \( -\ln(n+1) \).
This insight helps us see what the series approaches as n grows very large. Thus, dissecting a series into partial sums can reveal much about the nature of the series.

Exploring Logarithmic Properties

Logarithms are an essential tool in calculus and algebra. They transform multiplication into addition, division into subtraction, and powers into multiplications. At the heart of many problems is the need to understand and manipulate logarithmic expressions. Here’s how they work in the context of the series:

• We use the property: \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \).
• Applying this to \( \ln\left( \frac{n}{n+1} \right) \) transforms it into \( \ln(n) - \ln(n+1) \).
• This transformation is referred to as rewriting the terms to telescope, meaning many terms cancel each other when summed consecutively.

In our example, breaking logarithm expressions into simpler forms using properties allows simplification and helps identify patterns, enabling series evaluation.
Remember, understanding these properties can make evaluating, simplifying, and even solving logarithmic equations more straightforward.

Infinite Series and Divergence

The concept of an infinite series is when the series has an unending number of terms. Evaluating such series often involves looking at the limit of its partial sums as "n" approaches infinity. However, not all infinite series converge to a nice number; some diverge. Here's what happened in this exercise:

• The nth partial sum we discovered was \( S_n = -\ln(n+1) \).
• To evaluate an infinite series, we check the limit \( \lim_{n \to \infty} S_n \).
• As we let \( n \to \infty \), \( \ln(n+1) \) grows without bounds, meaning \( -\ln(n+1) \) goes to \(-\infty\).

This means our series does not settle to a finite value, leading us to conclude that the series diverges.
In simple terms, rather than approaching a single number, the series spreads out infinitely as n increases. Recognizing whether a series converges or diverges is a fundamental skill in mathematical analysis, and mastering it will aid in understanding more complex mathematical scenarios.