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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

The above problem came up in the now famous gAr series thread. Generally I stay away from that thread. It is for the members to work on and enjoy without me butting in.

I was asked for some help on this one and because this one has many interesting points I agreed. Of course before I attempt to answer I must do some ranting.

begin Rant():

I believe that this sort of problem is easily handled through the methods I have been going on and on about in this thread for more than a century now. Apparently I can foam at the mouth for as long as I like about how no one is reading any of it.

Why is that? Why is there not 1 million replies to each of these problems? Why do posters keep heading over to other forums to read non solutions full of homeomorphisms, iosomorphisms and more rings than a jeweler? Beats me!

End Rant:

Return(answer)

Here is how we can do it using the methods outlined here. These methods allow one to get exact answers to dificult problems using commonsense reasoning.

First thing we observe that as n gets larger so does k and so does k*n. That means those 3 constants a,b,c are going to be drowned out. We can reduce the problem to

and then to this.

This is easily handled by a CAS:

```
n = 300000000000000000000000;
NSum[1/Sqrt[k n + n^2], {k, 1, n}, WorkingPrecision -> 25]
```

the output is 0.8284271247461900976033770...

To show we are on the right track we do a little bit of experimenting. We choose three arbitrary values for a,b and c.

```
NSum[Sqrt[k n + n^2 + 31]/(
Sqrt[k n + n^2 + 2] Sqrt[k n + n^2 + 5]), {k, 1, n},
WorkingPrecision -> 25]
```

the output is 0.8284271247461900976033770...

We will get this for any a,b and c we choose.

Okay we have experimentally

what now? A PSLQ of course! We use one on the above constant and come up with:

We have a conjecture, a good one. We are done.

**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: 15,522

So,you hqve no definite proof of your answer.Ahhhhh... Well this will do I guess. But I would still like to do it by a hand method.

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: 86,636

Does a hand method automatically mean a proof?

**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: 15,522

If it is correct,yes.

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: 86,636

If it is long and difficult how do we verify its correctness?

**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: 15,522

How do we verify the correctness of your answer?

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: 86,636

We are sure of the amount of digits -1 that are provided. It is now possible to back engineeer that result. One method suggests itself right away. The point is, it is much better to have 25 digit conjecture than a 0 digit nothing.

But it is okay that you want to do it your way. I can now submit my answer to the series thread. When you get yours you can submit that also.

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: 15,522

Ok. I think I am gonna try it "my way",but even then I will use your result from computer math to help me. So it is not a waste in the end.

*Last edited by anonimnystefy (2012-04-21 23:59:40)*

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Do you have a suggestion for a hand method?

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
- Posts: 86,636

I am sorry, I do not.

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: 15,522

Okay. Thank,anyways,for this method here.

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
- Posts: 86,636

I think my earlier comments about a gAr problem apply here doubly.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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