Series Convergence: Explaining P>1 & P>0

[eyesontheball1]

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

 

[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.

 

[Ray Vickson]

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

 

[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)]

 

[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.

 

[eyesontheball1]

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

 

[Ray Vickson]

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
\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}

 

[eyesontheball1]

Can someone please elaborate on why my reasoning is false?

 

[eyesontheball1]

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

 

[Ray Vickson]

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).

 

[eyesontheball1]

I thought so.

 

[Ray Vickson]

I thought so.

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

 

[eyesontheball1]

Thank you for the help, Ray.

 

[Ray Vickson]

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
\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} \\<br /> = 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} \\<br /> = 1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots \pm \frac{1}{N^p} - 2 \sum_{n&gt;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!

 

[eyesontheball1]

Gotcha!

 

[haruspex]

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?