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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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#35 2012-09-02 21:05:52
- anonimnystefy
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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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón