Problem 37 - Thomas Calculus

Question

In Exercises \(35-40,\) find a formula for the \(n\) th partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}(\ln \sqrt{n+1}-\ln \sqrt{n}) $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Simplify the Expression for Partial Sum

The given series is \( \sum_{n=1}^{\infty} (\ln \sqrt{n+1} - \ln \sqrt{n}) \). Use the properties of logarithms to simplify each term:\[ \ln \sqrt{n+1} - \ln \sqrt{n} = \frac{1}{2} (\ln (n+1) - \ln n). \]This can be further simplified using the property \( \ln a - \ln b = \ln \frac{a}{b} \):\[ \frac{1}{2} \ln \frac{n+1}{n}. \]

Step 2: Express Partial Sum \(S_n\)

The partial sum \(S_n\) of the series up to the \(n\)th term is given by:\[ S_n = \sum_{k=1}^{n} \frac{1}{2} \ln \frac{k+1}{k}. \]Each term \(\ln \frac{k+1}{k}\) telescopes after the sum, so we have:\[ \frac{1}{2} \left( \ln \frac{2}{1} + \ln \frac{3}{2} + \ldots + \ln \frac{n+1}{n} \right). \]

Step 3: Identify the Telescoping Pattern

Notice that the sequence forms a telescoping series. All intermediate terms cancel out, leaving:\[ S_n = \frac{1}{2} \ln \frac{n+1}{1} = \frac{1}{2} \ln (n+1). \]

Step 4: Determine Convergence by Taking Limit

Examine the limit of the partial sum \( S_n \) as \( n \to \infty \):\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{2} \ln (n+1). \]As \( n \to \infty \), \( \ln(n+1) \to \infty \), so \( \frac{1}{2} \ln(n+1) \to \infty \) as well. The limit is infinity.

Step 5: Conclude the Series Behavior

Since the limit of the partial sum \( \lim_{n \to \infty} S_n = \infty \), the series diverges.

Key Concepts

Partial Sums

A partial sum is the sum of the first few terms of a series, stopping at a certain term. It is denoted as \( S_n \) where \( n \) represents the number of terms considered. Partial sums help in understanding the behavior of a series — whether it converges (approaches a fixed number) or diverges (grows without bound).
When dealing with series, especially infinite ones, it's crucial to explore the sequence of these partial sums. This means calculating \( S_1, S_2, S_3, \) and so on, and examining how \( S_n \) behaves as \( n \) increases indefinitely.

• If \( S_n \) approaches a specific number as \( n \) goes to infinity, the series converges to that number.
• If \( S_n \) grows indefinitely, the series diverges.

Partial sums are especially useful in proving series convergence using properties like telescoping, as they clearly reveal the series' behavior after intermediate terms cancel out.

Telescoping Series

A telescoping series is one where adjacent terms cancel each other out partially or completely, leaving behind a simple expression for the partial sum. This cancellation is key to simplifying complex series and recognizing convergence or divergence easier.
In this specific problem, the series \( \sum_{n=1}^{\infty} (\ln \sqrt{n+1} - \ln \sqrt{n}) \) is telescoping. After using logarithmic identities, the series' structure means that all intermediate terms between the first and last terms in the partial sum expression cancel out. This cancellation reduces the series down to a much simpler form:

• Only the first and last terms in each expression for \( \frac{k+1}{k} \) actually contribute to the sum.

For the question at hand, this process reveals that the partial sum \( S_n = \frac{1}{2} \ln(n+1) \), signifying that as new terms are added, simplification remains manageable because the majority of the terms cancel out.

Logarithmic Series

In mathematics, a logarithmic series involves terms that contain logarithmic functions, such as \( \ln(x) \). This presence of logarithms can significantly influence the behavior and simplification of a series.
Logarithmic properties, like \( \ln(a) - \ln(b) = \ln(\frac{a}{b}) \), help simplify expressions by reducing the number of terms. For instance, each term in our example transforms from \( \ln \sqrt{n+1} - \ln \sqrt{n} \) to \( \frac{1}{2} \ln \frac{n+1}{n} \), exploiting such identities.
This transformation not only reduces complexity but also aids in identifying patterns for telescoping and partial sums, as seen in our problem.

• It's essential to understand logarithmic properties to tackle problems like these effectively.
• Simplifying using logarithms can highlight convergence or divergence patterns in a series.

In this context, using logarithms helped us demonstrate how all intermediate terms cancel, thus providing a clearer pathway to assessing the behavior of the series.