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#26 2012-07-23 03:45:44

bobbym
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Re: a challenging problem for all

If you meant an expression for the closed form then it is obviously not possible. The sum will either equal a constant or 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.

#27 2012-07-23 03:47:03

anonimnystefy
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Re: a challenging problem for all

It is possible.


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

#28 2012-07-23 03:49:48

bobbym
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Re: a challenging problem for all

With a variable as the upper index yes. But with an infinity up there the sum is either a constant like 1 / 2 ( just picked as an example ) or ∞.


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.

#29 2012-07-23 03:52:02

anonimnystefy
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Re: a challenging problem for all

It can be an expression in terms of x.


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

#30 2012-07-23 07:01:52

bobbym
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Re: a challenging problem for all

Yes, it could.


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.

#31 2012-08-15 23:05:45

rajinikanth0602
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Re: a challenging problem for all

hi bobbym any information about my question

#32 2012-08-15 23:35:48

bobbym
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Re: a challenging problem for all

Hi rajinikanth0602;





By the ratio test:









By the ratio test the series diverges for all values of x.


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.

#33 2012-09-02 19:17:30

rajinikanth0602
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Re: a challenging problem for all

i think post reply ''19'' is not correct

#34 2012-09-02 19:18:23

bobbym
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Re: a challenging problem for all

Hi;

Did you see post #32?

i will challenge that i will give 7000 rupees for solving dis problem or any important information about that sequence or range of that sequence

In exchange for the rupees I would like your acknowledgement that the problem has been solved with the proof that is in post#32, or do you require more?

As for post #19. It may be correct, I do not know. Still it is a very nice piece of work by a good man.


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.

#35 2012-09-02 21:05:52

anonimnystefy
Real Member

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Re: a challenging problem for all

Hi rajinikanth0602

What problems do you see in post #19?

Last edited by anonimnystefy (2012-09-02 21:06:11)


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

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