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**mukesh****Member**- Registered: 2010-07-18
- Posts: 30

evaluate

lim n!/n^n

n->infinity

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,557

0.

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**mukesh****Member**- Registered: 2010-07-18
- Posts: 30

hw?i didnt undrstand

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Intuitively obvious that it approaches 0.

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

(capital pi symbol is for 'the product of' )

i/n ≤1 => the product is less than 1

as n approaches infinity 1/n approaches zero = the limit tends to zero

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Hi Bob

Is there somethingissing from the explanation?

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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi Stefy

Is there somethingissing from the explanation?

aybe. y posts often have things issing. But what would you like e to include?

I'd put in an 'm' if I could think of a suitable place for it.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,544

Typing issues, sorry.

The last sentence of your explanation doesn't follow from the rest.

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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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

Maybe you should use the fact that (1/n) is one of the terms of the product and all of the others are one or less.

Then use the fact that if you multiply by a number between zero and one it reduces a positive number or keeps

it the same when it is equal to one.

Then I think you can make Bob's deduction.

In a formal proof that needs to be written out correctly. If you are doing a pure maths related course, then I

should do the rest as an exercise, mukesh, if you just copy someone else's version then you won't understand it.

anonimystefy: I agree. Strictly speaking in a formal proof that probably isn't enough. Suppose the product sequence

converges to one with increasing density, as n inreases, all near to one. The product could converge to a higher number.

*Last edited by SteveB (2013-08-16 07:29:12)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,248

Hi;

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

I have shown that the product can be written

As 1/n tends to zero as n tends to infinity, I claim the product tends to zero times something finite and therefore to zero.

What's wrong with that?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

I agree with Bob. My only criticism was a lack of clarity of the argument. I knew what you meant and it is okay as

a proof provided:

(1) The result concerning (1/n) is proven earlier in the course and can be used by referring to it (it should be usually).

(2) The fraction has to be trapped below a certain positive constant. In this case it is obviously one or less.

Each term is between zero and one, including a term equaling one.

Proving bobbym's identities looks difficult to me, so if the question is an exercise in proving things then they might be

challenges to go on to if the proof of the first problem was too easy.

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