Question

Use the limit comparison test to determine whether each of the following series converges or diverges. $$ \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) $$

Short answer

The series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \) converges.

Step-by-step solution

Understand the Series

We are given the series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \). To determine if it converges or diverges, we'll use the limit comparison test.

Choose a Comparison Series

A good comparison series is \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a p-series with \( p = 2 \). We know this series converges because \( p > 1 \).

Consider the Limit of Ratios

Use the limit comparison test by considering the terms \( a_n = \ln \left(1+\frac{1}{n^{2}}\right) \) and \( b_n = \frac{1}{n^{2}} \). Compute the limit: \[L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\ln \left(1+\frac{1}{n^{2}}\right)}{\frac{1}{n^{2}}}\]

Simplify the Limit

For \( n \) large, use the approximation \( \ln(1+x) \approx x \) for small \( x \). Therefore, \( \ln \left(1+\frac{1}{n^{2}}\right) \approx \frac{1}{n^{2}} \). Substitute this into the limit to simplify:\[L = \lim_{n \to \infty} \frac{\frac{1}{n^{2}}}{\frac{1}{n^{2}}} = \lim_{n \to \infty} 1 = 1\]

Apply the Limit Comparison Test

The limit \( L = 1 \) is finite and positive. Thus, by the limit comparison test, since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, the original series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \) also converges.

Key concepts

Series Convergence

Series convergence is about deciding if the sum of the infinite series has a finite value or if it just keeps growing indefinitely. Understanding whether a series converges or diverges is crucial for properly analyzing sequences. Specifically, a series converges if the sequence of its partial sums approaches a finite limit. If it diverges, these partial sums do not approach any set value but continue to grow.When you want to determine if a series converges, there are different tests you can use. One common method is the **Limit Comparison Test**, which compares the terms of the series in question with those of a known series. If the comparison series is known to converge, and the ratio of the terms of the two series approaches a finite, specific number, then the original series also converges.For instance, in the original exercise, the limit comparison test was used to determine the convergence of the series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \) by comparing it with the convergent p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Because the limit of the ratio of the two series equals 1, it implies convergence of the original series.

P-Series

A P-Series is a special kind of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of a p-series depends crucially on the value of \( p \).

• If \( p > 1 \), the p-series converges. This means the sum of all the terms results in a finite number.
• If \( p \leq 1 \), the p-series diverges, and the terms add up to infinity as you go to an infinite number of terms.

In the exercise, a p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) with \( p = 2 \) was used as a comparison series. Since \( p = 2 \), which is greater than 1, this ensures convergence. Such characteristics of p-series were critical in confirming that \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \) also converges when using the limit comparison test.

Logarithm Approximation

Logarithm approximation is a technique used to simplify logarithmic expressions, especially when the arguments are small. The approximation that is often used is \( \ln(1+x) \approx x \) when \( x \) is close to zero. This is particularly useful when examining the convergence of a series which includes the logarithmic function.In the given series \( \sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n^{2}}\right) \), the logarithm approximation \( \ln \left(1+\frac{1}{n^{2}}\right) \approx \frac{1}{n^{2}} \), greatly simplifies calculations for large \( n \). It allows the original complex term to be easily compared to a simpler term \( \frac{1}{n^{2}} \), which was used in the limit comparison test. By approximating the logarithm function in this way, we can reveal the underlying structure of the series, leading to a straightforward application of convergence tests.