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**Bezoux****Member**- Registered: 2013-11-17
- Posts: 7

Hi, everyone.

I had an interesting problem in class today. Find

Now, it's fairly obvious that the answer is 0, the question is how to prove it.

I had an immediate suggestion, obviously if

,

then the answer is 0. So, we had to prove that.

By definition,

Since both and M are positive, I can raise them to the degree of n, which is when I get , which is obviously correct for n>= a certain n0.

Now, my teacher told me this was wrong and the class ended before he could elaborate on why, and the test is next class, which worries me a bit. I do a lot of my proofs this way.

What she said is that we need to prove , after which we can easily use the squeeze theorem to solve the problem. Unfortunately, I've had no success in proving that statement.

Can you point out where I was wrong in my solution, or help me prove the teacher's statement? I keep getting stuck at , but the teacher doesn't want us to use e (or infinite geometric series, which would make the proof fairly straightforward by using the binomial formula).

Thank you in advance, I know I'm asking a lot.

*Last edited by Bezoux (2014-01-23 06:35:27)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

You mean you can not use the well known fact that

**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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**Bezoux****Member**- Registered: 2013-11-17
- Posts: 7

Yes.

Is there a way to prove that that sequence is always lesser than 3 without e?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,315

Hi;

I am not getting anywhere with the restrictions that are placed on the problem. But starting with

perhaps you could replace the n! with Stirlings formula.

**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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