Alternating Series Test: Understanding the First Condition

Click For Summary
The discussion centers on the first condition of the Alternating Series Test, which requires that the nth term is positive and monotonically decreasing. Participants express confusion over the restrictiveness of this condition, questioning its necessity alongside the limit condition. An example is requested to illustrate an alternating series that meets the positivity and limit criteria but fails to be monotonically decreasing, leading to divergence. The provided example demonstrates how the positive terms can dominate, preventing convergence despite the negative terms. Ultimately, the interplay between positive and negative terms is crucial for determining the convergence of the series.
LumenPlacidum
Messages
40
Reaction score
0
The alternating series test contains two conditions for convergence.

The first condition is that the nth term (extracting the power of -1) is always positive and monotonically decreasing.

The second is that the limit of that nth term goes to 0 as n goes to infinity.


I've seen a proof for it, and I've even proved it myself (although some years ago). I don't really understand why the first condition is so restrictive. It seems like the always-positive part is all you need combined with the limit going to 0.

Can someone show me an example of an alternating series for which the terms (again, without the power of -1) are always positive and that have a limit of 0, but which is NOT monotonically decreasing and diverges because of it?
 
Physics news on Phys.org
<br /> 1,\; -1,\; \frac{1}{2},\; -\frac{1}{2^2},\; \frac{1}{3},\; -\frac{1}{3^2},\; \ldots<br />

The sum of the positive parts tends to push the series towards +\infty, and the sum of the negative parts tends to push the series towards -\pi^2/6, so the negative parts cannot cancel the positive parts sufficiently for the series to converge.
 
Great, thanks so much!
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K