Math Is Fun Forum

EMPhillips1989

Member

Registered: 2008-01-21

Posts: 40

sequences

can anyone prove that

please help!!

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: sequences

Don't you mean:

  ?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: sequences

Consider how the expression looks for a given n.  In expanded form it will look like this:

Notice that the every fraction is less than or equal to 1.  This tells us that for a given n, we know that n!/n^n <= 1/n, because 1/n is a member of the expanded expression and it is being multiplied by other fractions that are less than 1.  Then you point out that 1/n tends to 0 as n tends to infinity, and qed.

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Wrap it in bacon

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: sequences

While Dude's idea is correct, and is indeed the way that you should think of and solve the problem, many professors will wish you to be more rigorous.  If such is the case, then you will want to explicity prove:

Or perhaps this is easier to comprehend:

This should be a very simple induction proof.

Edit: And there is a little gotcha that I didn't see at first glance.  You will want to state that each term in the sequence is bounded below by 0 (i.e. positive).  While this seems obvious, it is very important as this argument would not work if the sequence was not.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."