Is dx ever truly equal to zero in integration theory?

[AdkinsJr]

Associativity doesn't work in series. The general property only works for absolute convergent series.

That was my first thought too. It's very funny how they just pull out that 1, you could just as easily show that it is equal to 2 if you wanted. The conclusion in the back of the book is that the series is divergent, which I think is equivalent to what you said.

 

[DeIdeal]

The conclusion in the back of the book is that the series is divergent, which I think is equivalent to what you said.

It's not, though. A series $\Sigma a_n$ is called absolutely convergent if $\Sigma |a_n|$ converges. A counterexample to show that this not being the case doesn't imply divergence (ie. a series doesn't have to be absolutely convergent to converge) is the alternating harmonic series ($a_n=\frac{(-1)^{n+1}}{n}$; $\Sigma_{n=1}^{\infty}a_n=\ln(2)$). It is not divergent, but as the harmonic series is, ($|a_n|=\frac{1}{n}$) it's not absolutely convergent either, and it's instead said that it converges conditionally.

In fact, I believe that Riemann proved a theorem ( I can't remember the exact name of the theorem) that essentially states that there exists a way to rearrange the terms of a conditionally convergent series so that the series can be made to diverge or to conditionally converge to any real number.

EDIT: It seems that it's called Riemann's Rearrangement Theorem.

 

[John Sovereign]

Think of the sum:

1+1+1+1+1+1+...=-1/2

This sum is only well defined for ζ(0), but we will assume that it is always true, so we multiply both sides by a real variable 'a'.

From which we obtain:

a+a+a+a+a+...=-a/2

Let a=0

0+0+0+0+...=-0/2
0+0+0+0+...=0

That is the first approach.

 

[John Sovereign]

The second approach would be using the infinite geometric series:

1+x+x^2+x^3+x^4+...=1/(1-x), for abs(x)<1

We can subtract 1 from each side:

x+x^2+x^3+x^4+...=x/(x-1)

Let x=0

0+0+0+0+...=0

 

[John Sovereign]

The last approach is by adding the sum term by term:

0, 0+0, 0+0+0, 0+0+0+0, ...

We can carry writing this forever, but the answer will never change, thus the limit won't change.

Thus 0+0+0+...=0

We can also try averaging the partial sums, but we'll still end up with 0.

 

[WWGD]

Oh God. What did I write I corrected it.

I just emailed it to your advisor. Please don't email my gross mistakes to anyone :).

 

[WWGD]

Re dx=0 , since I am not fully clear of the context, if you mean in the way it is used in integration theory, then the answer is no;
actually, for a _ fixed Riemann sum_, dx is defined as the least value $x_k -x_{k-1}$ , where the $x_k$ are part of a partition of an interval (integrals on unbounded interval are a separate issue), and it takes on a well-defined minimum value.
For the actual Riemann integral, you consider the _limit_ as $dx \rightarrow 0$ , bt dx is itself not 0, at least not so in this sense of the word.