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## #1 2012-04-16 08:15:54

bobbym

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### A tough limit

This problem came up in another thread. Let's see if it can be done using a CAS and some experimental techniques.

First we attempt to approximate the sum as accurately as we can:

with confidence in all the digits. Next a PSLQ was done to try to identify the number in terms of simple constants. This turned out to be fruitless. So the Euler Mclaurin formula was used next.

Basically the EMS is a formula that relates sums, integrals and derivatives together. It has many forms but the one we are interested in looks like this.

Often this form is of use for tough sums because it is often easier to integrate and differentiate. Plugging the above sum into that equation and asking mathematica to evaluate it produces a big mess. With some work you can get this out of it.

We can take the limit as n approaches infinity term by term and we are left with

so

and we are done.

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.

## #2 2012-04-16 09:19:43

anonimnystefy
Real Member

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### Re: A tough limit

Hi bobbym

I also used EMS,but I used the one from wiki,and it leads you to the answer in just 1 step. You have to calculate an integral and that's it.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #3 2012-04-16 09:25:21

bobbym

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### Re: A tough limit

So does mine, all the remaining are just to show that they are all zero. Only the integral is is used here. Most of the time that will not be the case.

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.

## #4 2012-04-16 09:30:09

anonimnystefy
Real Member

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### Re: A tough limit

But where did the rest of the infinity number of terms go?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #5 2012-04-16 09:31:26

bobbym

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### Re: A tough limit

I am assuming they all are zero when n approaches infinity.

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.

## #6 2012-04-16 09:39:14

anonimnystefy
Real Member

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### Re: A tough limit

Oh,ok. I think that method can be used on another sum in the thread.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #7 2012-04-16 10:27:17

bobbym

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### Re: A tough limit

It is a general purpose method light on rigor but heavt\y on results.

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.