Falling functions

  • Thread starter lukatwo
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  • #1
lukatwo
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Homework Statement



Hello, I was wondering if a falling function needs to fall on it's whole domain? I was solving a series with alternating sings. It says that if the series has alternating signs, that it converges if:
1) lim(n->0)an=0
2) an is a falling function
I was solving this series, and wasn't sure if the function was falling because the series starting from n=2
As you can see in this picture, this function goes ->0 for n->+inf, but starting from 2 it falls from -inf and goes to +inf, so it's not a fall?

Homework Equations



Similarly, (2*n+3)/(n^2) for n=[1,3]

The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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"Falling function"? Do you mean "decreasing function? If your question is just whether an "alternating series" converges when it is "eventually" decreasing, the answer is yes. The sum up to the point where the series begins decreasing is a finite number and has no effect on whether the series converges.
 
  • #3
lukatwo
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Yes, that is what I meant. English is my second language so math terminology in English is not my strong point. That was my question, and thanks for clearing it up!
 
  • #4
Ray Vickson
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Homework Statement



Hello, I was wondering if a falling function needs to fall on it's whole domain? I was solving a series with alternating sings. It says that if the series has alternating signs, that it converges if:
1) lim(n->0)an=0
2) an is a falling function
I was solving this series, and wasn't sure if the function was falling because the series starting from n=2
As you can see in this picture, this function goes ->0 for n->+inf, but starting from 2 it falls from -inf and goes to +inf, so it's not a fall?

Homework Equations



Similarly, (2*n+3)/(n^2) for n=[1,3]

The Attempt at a Solution


What you wrote above is wrong: you need to say that the a_n are alternating in sign, and that the absolute value |a_n| is decreasing to zero as n → ∞. (You did not mention the absolute value!) Furthermore, if you have a finite sum, the behaviour of the a_n is irrelevant, as has already been pointed out to you.
 
  • #5
lukatwo
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What you wrote above is wrong: you need to say that the a_n are alternating in sign, and that the absolute value |a_n| is decreasing to zero as n → ∞. (You did not mention the absolute value!) Furthermore, if you have a finite sum, the behaviour of the a_n is irrelevant, as has already been pointed out to you.

My textbook doesn't say it needs to be the absolute value, although it's under the absolute convergence chapter. When I think of it, it makes a lot of sense that it needs to be absolute convergence. But then again if we pair lim(n->inf)a_n=0 with a decreasing function I think we get the same thing.
 
  • #6
Ray Vickson
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My textbook doesn't say it needs to be the absolute value, although it's under the absolute convergence chapter. When I think of it, it makes a lot of sense that it needs to be absolute convergence. But then again if we pair lim(n->inf)a_n=0 with a decreasing function I think we get the same thing.

No: you can have alternating series that are convergent but NOT absolutely convergent! The simplest example is
[tex] S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln(2)[/tex]
but the 'absolute' series
[tex] 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \leftarrow \text{ diverges}[/tex]

Anyway, for a_n alternating in sign it makes no sense to speak of a_n decreasing; sometimes it increases and sometimes it decreases! For example, go from 1 to -1/2 (decrease); then go from -1/2 to +1/3 (increase), etc. You NEED the absolute value signs to make sense of the concept of decreasing'---no matter what your book does, or does not say.
 
  • #7
lukatwo
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That makes sense, but I probably should have said that I was wondering if it converges conditionally. Thanks for clearing things up!
 

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