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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

I don't mind having the discussion.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

Thanks gAr.

Can you post the accelerator Borwein found?

I am sure I showed it to you already. I know we discussed it. It is also useless for this sequence. I used Romberg and it worked well.

Want the code for Borwien's?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Yes. Is it the one with all the square roots and stuff?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

```
accelerate[n_]:=Module[{d,b,c,s},
d=(3+Sqrt[8])^n;
d=(d+1/d)/2;
b=-1;
c=-d;
s=0;
Table[c=b-c;s=s+c*a[k];b=(k+n)(k-n) b/((k+1/2)(k+1)),{k,0,n-1}];
s/d]
```

Remember how to use it?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

That's the one I was thinking of.

I set a[n] to be the array of the series terms. Then I do N[acc[number_of_terms],number_of_digits.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

I am pretty sure it is just for alternating sequences.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Yes, I know. It also has to start at 0.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

You can adjust the index to handle that. Anyways it is not useful on this problem.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Hi bobbym

You didn't tell me how we actually get the numerical answer here.

*Last edited by anonimnystefy (2013-12-11 10:57:23)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

Hi;

See post #527.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,525

Hm, isn't romberg for integrals?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,695

It is for sequences that are the result of numerical integrations on an integral. Actually it is a bunch of sequence accelerators.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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