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The proof of Rabbe's test

by quasar987
Tags: proof, rabbe, test
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quasar987
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Jun21-07, 12:22 PM
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1. The problem statement, all variables and given/known data
I was asked to prove Raabes test in the form that if

[tex]|a_{n+1}/a_{n}|<1-A/n[/tex]

for some A>1 and for n large enough, then the series converges absolutely.

After struggling for days (I work at a factory doing repetitive work so I can easily let my body work while I think about math ), I capitulated and peeked at the hint at the end of the book. It says, let

[tex]P_n=\prod_{k=1}^{n}\left(1-\frac{A}{k}\right)[/tex]

and show that

[tex]\ln(P_n)=-A\ln(n)+O(1)[/tex].

While I understand perfectly how this equality is the kernel of the proof, I can't seem to be able to demonstrate its truth. My best shot is...

[tex]\ln(P_n)=\sum_{k=1}^n\ln\left(1-\frac{A}{k}\right)\leq \sum_{k=1}^n\left(1-\frac{A}{k}\right)=n-A\sum_{k=1}^n\frac{1}{k}=n-A\ln(n)+O(1)[/tex]

The last equality is because the sequence [tex]\gamma_n=\sum_{k=1}^n\frac{1}{k}-\ln(n)[/tex] is decreasing and bounded below by 0 (and converges to Euler's cst [itex]\gamma[/itex]).
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NateTG
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Jun21-07, 12:50 PM
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Quote Quote by quasar987 View Post
While I understand perfectly how this equality is the kernel of the proof, I can't seem to be able to demonstrate its truth. My best shot is...

[tex]\sum_{k=1}^n\ln\left(1-\frac{A}{k}\right)\leq \sum_{k=1}^n\left(1-\frac{A}{k}\right)[/tex]
That seems like a poor choice of inequality -- the terms on the left are pretty close to zero, while the ones on the right are close to 1. I wonder if you could use the Taylor series:
[tex]\ln(1+x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}x^{n+1}[/tex]
to come up with something better?
quasar987
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Jun21-07, 01:09 PM
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I wonder...

I'd be surprised if that is the only way to go however. Cuz this is a problem from chapter 2, and Taylor's thm is in somthing like cahpter 6.

quasar987
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Jun22-07, 10:22 AM
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The proof of Rabbe's test

First, let me state correctly the version of Raabe's test I was asked to prove: "If

[tex]|a_{n+1}/a_{n}|\leq 1-A/n[/tex]

for some A>1 and for n large enough, then the series converges absolutely. And if

[tex]|a_{n+1}/a_{n}|\geq 1-1/n[/tex]

for n large enough, then the series diverges."

And I suppose that it is implied that if neither of these inequality hold for n large enough, then we cannot conclude.


Is this version of Raabe's test really equivalent to the one that says, "Let

[tex]L=\lim_{n\rightarrow\infty}n\left(1-\left|\frac{a_{n+1}}{a_n}\right|\right)[/tex]

If L>1, the series converges absolutely.
If L<1, the series diverges.
If L=1, we cannot conclude."

Or is this last one more general?

(And by the way, I think something's not right... cuz if we take the case [itex]|a_{n+1}/a_{n}|\geq 1-1/n[/itex], then the first test asserts convergence. But this equality also means that

[tex]n\left(1-\left|\frac{a_{n+1}}{a_n}\right|\right)\leq 1[/tex].

But in the case =1, the second test asserts that we cannot conclude. The two statements contradict each other, do they not?)
NateTG
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Jun22-07, 01:47 PM
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Quote Quote by quasar987 View Post
I'd be surprised if that is the only way to go however. Cuz this is a problem from chapter 2, and Taylor's thm is in somthing like cahpter 6.
Well, you can fudge it:
[tex]x \in (-\frac{1}{2},0)[/tex]
Gives you
[tex]1 < \frac{d}{dx} \ln(1+x) < 2 [/tex]
and
[tex]\ln 1= 0[/tex]
so on the same interval
[tex]x > \ln(1+x) > 2x[/tex]
Which is plenty strong for your needs.

You've got some issues with the inequalities in your expressions since both the conditions for convergence and divergence allow for equality.

You're not guaranteed that conditions are sufficient to guarantee that the limit exists so the inequalities are actually more general. Beyond that, it's quite easy to convert from one to the other using the notion of limit and some basic algebra:
[tex]\left| \frac{a_n}{a_{n+1}} \right | \leq 1 - \frac{A}{n}[/tex]
[tex]-\left|\frac{a_n}{a_{n+1}} \right | \geq \frac{A}{n} - 1[/tex]
[tex]1-\left| \frac{a_n}{a_{n+1}}\right| \geq \frac{A}{n}[/tex]
[tex]n\left(1-\left| \frac{a_n}{a_{n+1}}\right|\right) \geq A[/tex]
Now, the RHS is independant of [itex]n[/itex] so we can translate the 'sufficiently large' notion into a limit (provided the limit exists).
[tex]\lim_{n \rightarrow \infty} n\left(1-\left| \frac{a_n}{a_{n+1}}\right|\right) \geq A[/tex]
quasar987
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Jun22-07, 03:17 PM
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Quote Quote by NateTG View Post
You've got some issues with the inequalities in your expressions since both the conditions for convergence and divergence allow for equality.
I had a mistake in my post, sorry!

I edited it; the mistake is that the series will diverge if

[tex]|a_{n+1}/a_{n}|\geq 1-1/n[/tex]

for n large enough, and not if

[tex]|a_{n+1}/a_{n}|\geq 1-A/n[/tex]


Thanks a bunch for everything NateTG! <3
Hurkyl
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Jun22-07, 04:27 PM
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Rabbe's test (as stated) can be applied even when the ratio of terms doesn't converge. e.g.

1, 1/2, 1/8, 1/16, 1/64, 1/128, 1/512, 1/1024, ...
Enjolras1789
#8
Sep9-08, 10:09 AM
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You should use the proof of Kummer's test. Rabbe's test is a specific, less-general form of Kummer's test.


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