Series Convergence

  • #1

Homework Statement



Hi, everyone. I'd appreciate it if someone could explain something for me regarding the convergence of series. Thanks in advance![/B]

Homework Equations



In my calculus book, I'm given the following:

(1) - For p > 1, the sum from n=1 to infinity of n^-p converges.

(2) - For the sum from n=1 to infinity of [(-1)^(n+1)]*(n^-p), if lim of n^-p approaches 0 as n approaches infinity and if (n+1)^-p <= n^-p, then this alternating series converges. It's clear that this series converges if p > 0.

So we have two series, series (1), which converges whenever p > 1, and series (2), which converges whenever p > 0. What I don't understand is why exactly I'm wrong in the following reasoning:

Suppose p > 1. p > 1 =>

sum{n=1, infinity}{n^-p} converges and 1-2^(1-p) converges =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges, and

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity}{n^-p} - 2*sum{n=1, infinity}{(2n)^-p} =

sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges, but

sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0 =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0, and

1-2^(1-p) /= 0 whenever p /= 1 =>

sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)}/[1-2^(1-p)] converges whenever p > 0, p /= 1,

but we already know that sum{n=1, infinity}{n^-p} only converges for p s.t. p > 1, thus, we've arrived at a contradiction.

The Attempt at a Solution

 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
37,385
7,362
You started off with the assumption that p > 1, and made use of that. You cannot later in the argument deduce anything about the case of p <= 1.
Also, I couldn't follow what happened to the (2n)^-p term. It would be a lot easier to read if you take the trouble to use LaTeX.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



Hi, everyone. I'd appreciate it if someone could explain something for me regarding the convergence of series. Thanks in advance![/B]

Homework Equations



In my calculus book, I'm given the following:

(1) - For p > 1, the sum from n=1 to infinity of n^-p converges.

(2) - For the sum from n=1 to infinity of [(-1)^(n+1)]*(n^-p), if lim of n^-p approaches 0 as n approaches infinity and if (n+1)^-p <= n^-p, then this alternating series converges. It's clear that this series converges if p > 0.

So we have two series, series (1), which converges whenever p > 1, and series (2), which converges whenever p > 0. What I don't understand is why exactly I'm wrong in the following reasoning:

Suppose p > 1. p > 1 =>

sum{n=1, infinity}{n^-p} converges and 1-2^(1-p) converges =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges, and

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity}{n^-p} - 2*sum{n=1, infinity}{(2n)^-p} =

sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges, but

sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0 =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)} converges whenever p > 0, and

1-2^(1-p) /= 0 whenever p /= 1 =>

sum{n=1, infinity}{n^-p} = sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p)}/[1-2^(1-p)] converges whenever p > 0, p /= 1,

but we already know that sum{n=1, infinity}{n^-p} only converges for p s.t. p > 1, thus, we've arrived at a contradiction.

The Attempt at a Solution


In (2), do you mean
[tex] \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^p}, \; p > 0 \, ?[/tex]
Yes, indeed, it is convergent. Have you heard of the "alternating series test"? See, eg., http://en.wikipedia.org/wiki/Alternating_series_test
 
  • #4
I apologize for not using LateX. I was a bit short on time when I made the post. What if I instead argued as follows:

Suppose p > 0, p /=1.

p > 0, p /=1 => sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) converges, and sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) = [1-2^(1-p)]*sum{n=1, infinity}{n^-p} =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges => [1-2^(1-p)]
 
  • #5
I apologize for not using LateX. I was a bit short on time when I made the post. What if I instead argued as follows:

Suppose p > 0, p /=1.

p > 0, p /=1 => sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) converges, and sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) = [1-2^(1-p)]*sum{n=1, infinity}{n^-p} =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges => [1-2^(1-p)] converges and sum{n=1, infinity}{n^-p} converges => sum{n=1, infinity}{n^-p} converges whenever p > 0, p /= 1, but we've already been given that sum{n=1, infinity}{n^-p} converges only for p > 1.
 
  • #6
Also, please ignore the first of the two replies above.
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I apologize for not using LateX. I was a bit short on time when I made the post. What if I instead argued as follows:

Suppose p > 0, p /=1.

p > 0, p /=1 => sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) converges, and sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) = [1-2^(1-p)]*sum{n=1, infinity}{n^-p} =>

[1-2^(1-p)]*sum{n=1, infinity}{n^-p} converges => [1-2^(1-p)] converges and sum{n=1, infinity}{n^-p} converges => sum{n=1, infinity}{n^-p} converges whenever p > 0, p /= 1, but we've already been given that sum{n=1, infinity}{n^-p} converges only for p > 1.

You say
[tex] \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^p} = \left(1-2^{1-p} \right) \sum_{n=1}^{\infty} \frac{1}{n^p} \;\Longleftarrow \;\text{false reasonng} [/tex]
 
Last edited:
  • #8
Can someone please elaborate on why my reasoning is false?
 
  • #9
Also, does the flaw in my reasoning have something to do with the Reimann rearrangement theorem?
 
  • #10
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Also, does the flaw in my reasoning have something to do with the Reimann rearrangement theorem?

As far as I can see you did not do any "reasoning" at all, but just wrote down some things without much justification.

That said: what you wrote down appears to be true for integers p = 2,3,4, ... ! It may also be true for non-integer p > 1, but that is harder to justify. Maple can evaluate the sums numerically. Even to 40-digit accuracy or more, Maple gets the same numbers on both sides for integer p > 1, but can only match about the first 10 or 11 digits when p > 1 is fractional (with different levels of accuracy for different values of p).
 
  • #12
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I thought so.

Nevertheless, your "reasoning" had no substance; you really need to do things carefully and convincingly. Otherwise, nobody will believe you.
 
  • #14
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Thank you for the help, Ray.

OK, I see how to fix it up your basic argument. Let ##p>1##. Then, for finite integer ##N > 0## we have
[tex] \left(1-2^{1-p} \right) \sum_{n=1}^N \frac{1}{n^p} = \sum_{n=1}^N \frac{1}{n^p} - 2 \sum_{n=1}^N \frac{1}{(2n)^p} \\
= 1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{N^p} - \frac{2}{2^p} - \frac{2}{4^p} - \frac{2}{6^p} - \cdots -\frac{2}{(2N)^p} \\
= 1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots \pm \frac{1}{N^p} - 2 \sum_{n>N/2, n \leq N} \frac{1}{(2n)^p}.[/tex]
Since ##p > 1## the "error" term ##2 \sum_{n>N/2, n \leq N} \frac{1}{(2n)^p} \to 0## as ##N \to \infty##, so we end up with your result
[tex] \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^p} = \left(1-2^{1-p} \right) \sum_{n=1}^{\infty} \frac{1}{n^p}[/tex]

Well done!
 
Last edited:
  • #16
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
37,385
7,362
It may also be true for non-integer p > 1, but that is harder to justify.
Isn't it fairly straightforward for all p > 1, using fact (1) given in the OP?
 

Related Threads on Series Convergence

  • Last Post
Replies
5
Views
799
  • Last Post
Replies
2
Views
909
  • Last Post
Replies
8
Views
726
  • Last Post
Replies
3
Views
895
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
774
  • Last Post
Replies
1
Views
870
  • Last Post
Replies
6
Views
1K
Top