Solving Abs Converge Issue with Ratio Test

  • Context: MHB 
  • Thread starter Thread starter Dustinsfl
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the application of the Ratio Test to determine the convergence of the Fourier series for the function \(f(\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}\sin n\theta\). The user mistakenly concludes absolute convergence based on the Ratio Test, which yields a limit of 1, indicating that the test is inconclusive. The correct interpretation is that the Ratio Test cannot confirm convergence when the limit equals 1, and the user confuses this with the Limit Comparison Test, which has different criteria for establishing convergence.

PREREQUISITES
  • Understanding of Fourier series and their convergence properties
  • Familiarity with the Ratio Test for series convergence
  • Knowledge of the Limit Comparison Test
  • Basic calculus, particularly limits and series
NEXT STEPS
  • Study the conditions under which the Ratio Test is inconclusive
  • Learn about the Limit Comparison Test and its applications
  • Explore absolute and conditional convergence in series
  • Investigate other convergence tests such as the Root Test and Integral Test
USEFUL FOR

Mathematicians, students studying calculus or real analysis, and anyone interested in series convergence and Fourier analysis.

Dustinsfl
Messages
2,217
Reaction score
5
The book says this isn't absolutely convergent but I keep getting it is by the ratio test. What is wrong?

The Fourier series for $f$ is $f(\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}\sin n\theta$.
Then
$$
-\sum\limits_{n = 1}^{\infty}\left|\frac{(-1)^{n + 1}}{n}\right|.
$$
By the ratio test, we have
\begin{alignat*}{3}
\lim_{n\to\infty}\left|\frac{(-1)^{n + 2}n}{(-1)^{n + 1}(n + 1)}\right| & = & \lim_{n\to\infty}\left|\frac{-n}{n + 1}\right|\\
& = & \lim_{n\to\infty}|-1|\frac{n}{n + 1}\\
& = & 1 < \infty
\end{alignat*}

I solved this problem in another manner but shouldn't I be able to get the same answer using the Ratio Test?
 
Last edited:
Physics news on Phys.org
dwsmith said:
I solved this problem in another manner but shouldn't I be able to get the same answer using the Ratio Test?
The ratio test is inconclusive if the limit \(L = 1\). You are perhaps confusing this with the limit comparison test, which only requires the limit to exist and be nonzero in order to establish convergence or divergence.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K