Question

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

Short answer

The interval of convergence is \([-1, 1)\).

Step-by-step solution

- Identify the General Term

The given series is \(\textstyle\sum_{n=1}^{\infty} \frac{x^{n}}{\ln (n+1)}\). Identify the general term, \(a_n\), which in this case is \(\frac{x^n}{\ln(n+1)}\).

- Apply the Ratio Test

The Ratio Test can help find the interval of convergence. Consider \(\textstyle\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\): \[\textstyle\lim_{n \to \infty} \left|\frac{\frac{x^{n+1}}{\ln(n+2)}}{\frac{x^n}{\ln(n+1)}}\right| = \lim_{n \to \infty} \left|x \cdot \frac{\ln(n+1)}{\ln(n+2)}\right| = |x| \lim_{n \to \infty} \[\frac{\ln(n+1)}{\ln(n+2)}\] = |x| \cdot 1 = |x|\].

- Determine Convergence from Ratio Test

By the Ratio Test, the series converges if \[|x| < 1\] and diverges if \[|x| > 1\]. Therefore, the open interval of convergence is \((-1, 1)\).

- Test Endpoints x = -1 and x = 1

Test the series at the endpoints.\At \[x = -1\]: The series becomes \(\textstyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)}\), which is an alternating series. Since \(\textstyle\frac{1}{\ln(n+1)} \to 0\) as \(\textstyle n \to \infty\), the series converges by the Alternating Series Test.\At \[x = 1\]: The series becomes \(\textstyle\sum_{n=1}^{\infty} \frac{1}{\ln(n+1)}\), which diverges by the Integral Test as \(\textstyle\int_1^{\infty} \frac{1}{\ln(x+1)} \text{dx}\) diverges.

- Combine Results

Combine the results from the Ratio Test and the endpoint tests. The interval of convergence is \([-1, 1)\).

Key concepts

Ratio Test

The Ratio Test is a common method for determining the convergence of a series. The Ratio Test involves taking 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 is used as follows: \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L\]. If \[L < 1\], the series converges. If \[L > 1\], the series diverges. If \[L = 1\], the test is inconclusive.

In our problem, we applied the Ratio Test to the series \(\textstyle \sum_{n=1}^{\infty} \frac{x^n}{\ln(n+1)}\). By simplifying the limit, we found that \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = |x|\]. This tells us that the series converges for \[|x| < 1\], giving an interval of \((-1, 1)\).

Alternating Series Test

The Alternating Series Test is useful for determining the convergence of series whose terms alternate in sign. Consider a series of the form \(\textstyle\sum(-1)^n b_n\), where \[b_n > 0\]. For the Alternating Series Test:

• The series \[b_n\] is decreasing, and
• \[b_n \to 0\] as \[n \to \infty\].

If both conditions are met, the series converges.

In the given exercise, testing the endpoints included checking for convergence of \(\textstyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)}\). Here, \(\textstyle \frac{1}{\ln(n+1)}\) decreases and goes to 0 as \(\textstyle n \to \infty\). Thus, by the Alternating Series Test, the series converges when \[x = -1\].

Integral Test

The Integral Test relates the convergence of a series to the convergence of an integral. For a series \(\textstyle\sum_{n=1}^{\infty} a_n\), if \[a_n\] is positive, continuous, and decreasing, we can use the Integral Test by comparing the series to an improper integral. The series converges if and only if \[\int_1^{\infty} f(x) \text{dx}\] converges, where \[f(n) = a_n\].

For \[x = 1\] in our problem, the series \(\textstyle \sum_{n=1}^{\infty} \frac{1}{\ln(n+1)}\) was evaluated using the Integral Test. We looked at \[\int_1^{\infty} \frac{1}{\ln(x+1)} \text{dx}\], which diverges. This shows that the series also diverges when \[x = 1\].

Analysis of Series

Understanding the convergence and divergence of series is crucial in calculus. Each test provides a different approach to analyzing series:

• The Ratio Test is very versatile and often simplifies to a quick answer.
• The Alternating Series Test relies on behavior of alternating terms and is simpler to check when applicable.
• The Integral Test forms a bridge between understanding the sum of a series and integrals, often being useful for series with positive terms.

Combining these tests helps determine intervals of convergence. For \(\textstyle \sum_{n=1}^{\infty} \frac{x^n}{\ln(n+1)}\), the interval was found to be \([-1, 1)\) by using Ratio Test and end-point analysis with Alternating Series Test and Integral Test. This comprehensive understanding of series behaviors ensures a robust approach to many calculus problems.