Question

Find the interval of convergence, including end-point tests: \(\sum_{n=1}^{\infty} \frac{x^{n}}{\ln (n+1)}\)

Short answer

Interval of convergence: \(\textstyle -1 \leq x < 1\).

Step-by-step solution

- Apply the Ratio Test

To find the interval of convergence, start by applying the Ratio Test. The Ratio Test states that for a series \(\textstyle\sum_{n=1}^{\infty} a_n\), you should calculate \(\textstyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). Let's take \(\textstyle a_n = \frac{x^n}{\ln(n+1)}\).Compute: \(\textstyle \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{x^{n+1}}{\ln(n+2)}}{\frac{x^n}{\ln(n+1)}} \right| = \left| x \frac{\ln(n+1)}{\ln(n+2)} \right|\).

- Simplify the Ratio

Simplify the expression inside the limit: \(\textstyle \left| \frac{a_{n+1}}{a_n} \right| = \left| x \right| \cdot \frac{\ln(n+1)}{\ln(n+2)}\).Now evaluate the limit: \(\textstyle \lim_{n \to \infty} \left| x \right| \cdot \frac{\ln(n+1)}{\ln(n+2)} = \left| x \right| \cdot \lim_{n \to \infty} \frac{\ln(n+1)}{\ln(n+2)} = \left| x \right|\).The limit here is \(\textstyle \left| x \right| \).

- Apply the Ratio Test Result

According to the Ratio Test, the series converges if this limit is less than 1. Therefore, solve: \(\textstyle \left| x \right| < 1\).This indicates that the interval of convergence is \(\textstyle -1 < x < 1\).

- Test Endpoint: x = 1

Substitute \(\textstyle x = 1 \) into the original series: \(\textstyle \sum_{n=1}^{\infty} \frac{1^n}{\ln(n+1)} = \sum_{n=1}^{\infty} \frac{1}{\ln(n+1)}\).To determine convergence, use the integral test. The integral \(\textstyle \int \frac{1}{\ln(x+1)}dx\) diverges, thus the series at \(\textstyle x = 1 \) diverges.

- Test Endpoint: x = -1

Substitute \(\textstyle x = -1 \) into the original series: \(\textstyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)}\).This is an alternating series. Use the Alternating Series Test. For this test, \(\textstyle \frac{1}{\ln(n+1)} \) is positive, decreasing, and tends to 0 as \(\textstyle n \to \infty \). Thus, the series converges at \(\textstyle x = -1\).

- State the Interval of Convergence

Combine the results. The series converges for \(\textstyle -1 \leq x < 1\).

Key concepts

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of a series. It involves computing the limit of the absolute value of the ratio of consecutive terms. For a series \(\textstyle\sum_{n=1}^{\infty} a_n\), the Ratio Test requires you to find: \[\textstyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]. If this limit is less than 1, the series converges absolutely.

In the given exercise, we start by setting \(\textstyle a_n = \frac{x^n}{\ln(n+1)}\). After computing \(\textstyle \left| \frac{a_{n+1}}{a_n} \right| \), we simplify it to \(\textstyle \left| x \right| \cdot \frac{\ln(n+1)}{\ln(n+2)} \). By taking the limit as \(\textstyle n \to \infty\), we find that it equals \(\textstyle \left| x \right| \).

According to the Ratio Test, since \(\textstyle \left| x \right| < 1\), the series converges within the interval \(\textstyle -1 < x < 1\).

Alternating Series Test

The Alternating Series Test (also known as the Leibniz Test) is used to determine the convergence of series that alternate in sign. For a series of the form \(\textstyle\sum_{n=1}^{\infty} (-1)^{n} b_n \), where \(\textstyle b_n \) is a positive, decreasing sequence that approaches 0, the test confirms convergence if:

• \(\textstyle b_{n+1} \) is less than or equal to \(\textstyle b_n \)
• \(\textstyle \lim_{n \to \infty} b_n = 0 \)

In the problem, when \(\textstyle x = -1\), the series becomes \(\textstyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)} \). We need to ensure that \(\textstyle \frac{1}{\ln(n+1)}\) is a positive, decreasing term that approaches 0 as \(\textstyle n \to \infty \). By checking these conditions, we confirm the series converges, indicating convergence at \(\textstyle x = -1 \).

Integral Test

The Integral Test is another method used to determine series convergence by comparing it to an improper integral. Consider a series \(\textstyle\sum_{n=1}^{\infty} a_n \). If \(\textstyle a_n = f(n)\), where \(\textstyle f(x) \textstyle \ge \ 0\) is a continuous, decreasing function, then the series converges if and only if the integral \(\textstyle \int_{1}^{\infty} f(x) dx \) converges.

To apply to the exercise, when \(\textstyle x = 1\), the series transforms to \(\textstyle \sum_{n=1}^\infty \frac{1}{\ln(n+1)}\). Evaluating the integral \(\textstyle \int \frac{1}{\ln(x+1)}dx \), we find that it diverges, suggesting the series does not converge at \(\textstyle x = 1\).

Series Convergence

Series convergence refers to whether the sum of the sequence terms approaches a finite value as the number of terms grows indefinitely. For the given series \(\textstyle\sum_{n=1}^{\infty} \frac{x^n}{\ln(n+1)}\), we combined the Ratio Test and the endpoint evaluations using the Alternating Series Test and Integral Test to find the interval of convergence. By performing these tests:

• The Ratio Test gave \(\textstyle -1 < x < 1 \)
• The Integral Test showed that the series diverges at \(\textstyle x = 1 \)
• The Alternating Series Test demonstrated convergence at \(\textstyle x = -1 \)

We conclude the interval of convergence for the series is \(\textstyle -1 \le x < 1\).