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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

How did you get that?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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http://mathworld.wolfram.com/RiemannZetaFunction.html

Formula 19 - 20

But the important thing is why is it better?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Because it's alternating, of course. RRA is not working too fast, though. I cannot evaluate it quickly...

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
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Alternating series can be hard to accelerate too! Heard of the converger?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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No. What is it?

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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
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```
acc[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]
```

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

How does it work?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I knew you were going to say that...

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

And you do not know?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I knew you were going to ask that.

First, it is for alternating series only. So enter

a[k_]:=1 / ( k + 1 ) then

N[acc[250],200]

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi

I am getting 0.69314718055994530941723212145817656807550013436025525412068000949339362196969471560586332699641868754200148102057068573368552023575813055703267075163507596193072757082837143519030703862389167347122161...

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**bobbym****Administrator**- From: Bumpkinland
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Now enter:

N[Log[2],200]

what do you notice?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Wow, 196 digits!

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**bobbym****Administrator**- From: Bumpkinland
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No what series you just did?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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The converger always starts at 0.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Same series.

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**bobbym****Administrator**- From: Bumpkinland
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Yes, but it will affect what you enter into a[k_]

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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That is true, but you asked what the series was.

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**bobbym****Administrator**- From: Bumpkinland
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Yes, but I meant in the form that the converger can use.

1) The series must be alternating.

2) It must start at 0.

The function acc has an input and a global variable a. These must be in the precise form required. That is why I did the transformation on the zeta series.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

No, no, I got that it must start at 0. I was just saying that what I wrote and what you wrote are the same series.

So, you've got no idea how and why it works?

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**bobbym****Administrator**- From: Bumpkinland
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None at all. Did you see how it works on zeta(3)?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

Haven't tried. I will do it tomorrow.

What is tthe algorithm based on?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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The converger? I do not know,

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,184

Too bad.

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

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