Problem 42 - Single Variable Calculus

Question

Determine whether the given series converges or diverges. If it converges, find its sum. \(\sum_{n=1}^{\infty} \ln \left(\frac{n}{n+1}\right)\)

Step-by-Step Solution

Answer

The given series \(\sum_{n=1}^{\infty} \ln \left(\frac{n}{n+1}\right)\) can be rewritten using logarithmic properties as \(\sum_{n=1}^{\infty} (\ln(n) - \ln(n+1))\). Observing the partial sums, we notice a telescoping pattern, with terms cancelling out each other. The remaining terms are \(\ln(1) - \ln(n+1)\). As \(n\) approaches infinity, \(\ln(n+1)\) also approaches infinity, causing the series to diverge.

Step 1: Simplify the terms using properties of logarithms

Use the logarithmic property \(\ln(a) - \ln(b) = \ln(\frac{a}{b})\) to simplify the terms: \(\ln\left(\frac{n}{n+1}\right) = \ln(n) - \ln(n+1)\) Now the series can be rewritten as: \(\sum_{n=1}^{\infty} (\ln(n) - \ln(n+1))\)

Step 2: Observe the series for patterns

Now let's observe the partial sums of the series: \(S_1 = \ln(1) - \ln(2)\) \(S_2 = (\ln(1) - \ln(2)) + (\ln(2) - \ln(3))\) \(S_3 = (\ln(1) - \ln(2)) + (\ln(2) - \ln(3)) + (\ln(3) - \ln(4))\)

Step 3: Identify cancellation

In each partial sum, the negative terms cancel out the previous positive terms. This pattern is an example of telescoping series. The only terms that don't cancel out are the first and the last term of the partial sum: \(S_n = \ln(1) - \ln(n+1)\) As \(n\) approaches infinity, the logarithm function \(\ln(n+1)\) also approaches infinity.

Step 4: Determine convergence

Since the logarithm function increases without bound as \(n\) approach infinity, the series diverges: \(\lim_{n \to \infty} S_n = \lim_{n \to \infty} (\ln(1) - \ln(n+1)) = -\infty\)

Key Concepts

Logarithmic Series

The logarithmic series involves expressions where logarithms are used to form the terms of a series. A common form is using the property of logarithms to simplify expressions before summing them. Consider the series:\[\sum_{n=1}^{\infty} \ln \left(\frac{n}{n+1}\right)\]Utilizing the fundamental logarithmic property \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), each term can be rewritten, simplifying the series:\[\ln\left(\frac{n}{n+1}\right) = \ln(n) - \ln(n+1)\]By expressing the terms in this manner, recognizing patterns and evaluating convergence or divergence of the series becomes more approachable. This simplification exposes other mathematical structures or properties in a series, which might not be immediate otherwise.

Telescoping Series

A telescoping series is characterized by the cancellation of consecutive terms, revealing simplified results. The given series:\[\sum_{n=1}^{\infty} (\ln(n) - \ln(n+1))\]ensures that sequential terms effectively nullify each other. Here's how the cancellation works:

• Consider the partial sums: \( S_1 = \ln(1) - \ln(2) \)
• Then, \( S_2 = (\ln(1) - \ln(2)) + (\ln(2) - \ln(3)) \)
• Now, \( S_3 = (\ln(1) - \ln(2)) + (\ln(2) - \ln(3)) + (\ln(3) - \ln(4)) \)

Effectively, the middle terms cancel each other out. What's left in the partial sum \( S_n \) indicates that only the first and last terms survive:\[S_n = \ln(1) - \ln(n+1)\]This makes telescoping series a powerful tool for understanding complex series, as simplifications occur naturally through the pattern of the series, often reducing the complexity of calculating high sums to just a few terms.

Partial Sums

Analyzing partial sums helps determine the behavior of a series as more terms are added. For the series in question:\[S_n = \ln(1) - \ln(n+1)\]The nature of partial sums directly informs about the convergence or divergence of the series. Initially:

• \( S_1 = \ln(1) - \ln(2) \)
• \( S_2 = \ln(1) - \ln(3) \)
• \( S_3 = \ln(1) - \ln(4) \)

Notice how each partial sum \(S_n\) formula simplifies as more terms are added. The pattern emerges of cancellation due to the structure derived from the original series' terms. Crucially, as \( n \to \infty \), \( \ln(n+1) \to \infty \), indicating that:\[\lim_{n \to \infty} S_n = -\infty\]This divergent behavior, demonstrated through partial sums, reveals the nature of the series effectively.