Series Convergence

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1. Sep 28, 2014

eyesontheball1

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. The attempt at a solution

2. Sep 28, 2014

haruspex

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. Sep 29, 2014

Ray Vickson

In (2), do you mean
$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^p}, \; p > 0 \, ?$$
Yes, indeed, it is convergent. Have you heard of the "alternating series test"? See, eg., http://en.wikipedia.org/wiki/Alternating_series_test

4. Sep 29, 2014

eyesontheball1

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. Sep 29, 2014

eyesontheball1

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. Sep 29, 2014

eyesontheball1

Also, please ignore the first of the two replies above.

7. Sep 29, 2014

Ray Vickson

You say
$$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^p} = \left(1-2^{1-p} \right) \sum_{n=1}^{\infty} \frac{1}{n^p} \;\Longleftarrow \;\text{false reasonng}$$

Last edited: Oct 4, 2014
8. Oct 3, 2014

eyesontheball1

Can someone please elaborate on why my reasoning is false?

9. Oct 3, 2014

eyesontheball1

Also, does the flaw in my reasoning have something to do with the Reimann rearrangement theorem?

10. Oct 4, 2014

Ray Vickson

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).

11. Oct 4, 2014

eyesontheball1

I thought so.

12. Oct 4, 2014

Ray Vickson

Nevertheless, your "reasoning" had no substance; you really need to do things carefully and convincingly. Otherwise, nobody will believe you.

13. Oct 4, 2014

eyesontheball1

Thank you for the help, Ray.

14. Oct 4, 2014

Ray Vickson

OK, I see how to fix it up your basic argument. Let $p>1$. Then, for finite integer $N > 0$ we have
$$\left(1-2^{1-p} \right) \sum_{n=1}^N \frac{1}{n^p} = \sum_{n=1}^N \frac{1}{n^p} - 2 \sum_{n=1}^N \frac{1}{(2n)^p} \\ = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{N^p} - \frac{2}{2^p} - \frac{2}{4^p} - \frac{2}{6^p} - \cdots -\frac{2}{(2N)^p} \\ = 1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots \pm \frac{1}{N^p} - 2 \sum_{n>N/2, n \leq N} \frac{1}{(2n)^p}.$$
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
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^p} = \left(1-2^{1-p} \right) \sum_{n=1}^{\infty} \frac{1}{n^p}$$

Well done!

Last edited: Oct 4, 2014
15. Oct 4, 2014

eyesontheball1

Gotcha!

16. Oct 5, 2014

haruspex

Isn't it fairly straightforward for all p > 1, using fact (1) given in the OP?