The Alternating Dance
An Alternating Series is one where the terms switch signs, like \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \). Unlike positive series which just grow, alternating series "dance" back and forth around the limit.
Alternating Series Test (Leibniz Test)
A series \(\sum (-1)^{n+1} a_n\) (with \(a_n \gt 0\)) converges if:
- The terms are decreasing: \(a_{n+1} \le a_n\)
- The limit is zero: \(\lim_{n \to \infty} a_n = 0\)
The Integral Test
We can compare a series \(\sum a_n\) to an improper integral \(\int_1^\infty f(x) dx\). If the function \(f(x)\) is positive, decreasing, and continuous, then the series and the integral behave the same way: both converge or both diverge.
Visual Comparison
See how the series sum relates to the area under the curve.
- Left Sum > Integral
- Right Sum < Integral
Since the integral is finite (area is bounded), the series must also be finite.
Practice Problems
True/False
1. If \(\lim_{n \to \infty} a_n = 0\), then the alternating series \(\sum (-1)^n a_n\) converges.
False. The terms \(a_n\) must also be decreasing (eventually). For example, if \(a_n\) oscillates, the series might diverge.
2. The Integral Test can be used to determine if \(\sum \frac{\sin(n)}{n}\) converges.
False. The Integral Test requires the function \(f(x)\) to be positive, but \(\sin(x)\) oscillates between positive and negative values.
3. If \(\sum |a_n|\) converges, then \(\sum a_n\) converges.
True. This is the definition of absolute convergence implying convergence.
Fill in the Blank
1. For the Alternating Series Test, we need \(a_{n+1} \le a_n\) and \(\lim_{n \to \infty} a_n = \) ______.
0.
2. To use the Integral Test on \(\sum a_n\), we consider \(f(x)\) such that \(f(n) = a_n\). The function \(f(x)\) must be continuous, positive, and ______.
Decreasing.
3. The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) is called the ______ series.
Alternating Harmonic.
Full Problems
1. Determine if \(\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\) converges or diverges.
Converges. This is an alternating series with \(a_n = \frac{1}{\sqrt{n}}\). Since \(a_n\) is decreasing and \(\lim a_n = 0\), it converges by the Alternating Series Test.
2. Use the Integral Test to determine the convergence of \(\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}\).
Let \(f(x) = \frac{1}{x^2+1}\). This is continuous, positive, and decreasing. \(\int_1^\infty \frac{1}{x^2+1} dx = \lim_{t \to \infty} [\arctan(x)]_1^t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\). Since the integral converges, the series converges.
3. Does \(\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}\) converge absolutely, conditionally, or diverge?
Conditionally Convergent. It converges by AST (terms decrease to 0). However, \(\sum |\frac{(-1)^n n}{n^2+1}| = \sum \frac{n}{n^2+1}\) behaves like \(\sum \frac{1}{n}\) (Limit Comparison Test), which diverges. Thus, it does not converge absolutely.
4. Estimate the error in approximating \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}\) by the sum of the first 4 terms.
For an alternating series satisfying the AST conditions, the error \(|R_n| \le a_{n+1}\). Here, \(n=4\), so \(|R_4| \le a_5 = \frac{1}{5^3} = \frac{1}{125} = 0.008\).