Infimum of the (1+1/n)^n

[viswanathan]

I want to show that the infimum of the set of all numbers of the form [MATH]\bigg(1+\frac{1}{n}\bigg)^n[/MATH] where n=1,2,3,... is 2.

I believe I can prove this if I can show that
[MATH]\bigg(1+\frac{1}{n+1}\bigg)^{n+1} > \bigg(1+\frac{1}{n}\bigg)^n[/MATH]
But I don't know how to show this is true.
Someone please help.

 

[Deleted member 4993]

I want to show that the infimum of the set of all numbers of the form [MATH]\bigg(1+\frac{1}{n}\bigg)^n[/MATH] where n=1,2,3,... is 2.

I believe I can prove this if I can show that
[MATH]\bigg(1+\frac{1}{n+1}\bigg)^{n+1} > \bigg(1+\frac{1}{n}\bigg)^n[/MATH]
But I don't know how to show this is true.
Someone please help.

If each term of the sequence an, then you want to show:

\(\displaystyle \frac{a_{n+1}}{a_n} \ \gt \ 1 \)

continue....

 

[JeffM]

I want to show that the infimum of the set of all numbers of the form [MATH]\bigg(1+\frac{1}{n}\bigg)^n[/MATH] where n=1,2,3,... is 2.

I believe I can prove this if I can show that
[MATH]\bigg(1+\frac{1}{n+1}\bigg)^{n+1} > \bigg(1+\frac{1}{n}\bigg)^n[/MATH]
But I don't know how to show this is true.
Someone please help.

Did you think about a proof by induction?

 

[Deleted member 4993]

The problem statement is incorrect.

The infimum supremum should be e (Euler's number). However, if you want to proceed in your way: ....................... edited

If each term of the sequence a_n, then you want to show:

\(\displaystyle \frac{a_{n+1}}{a_n} \ \gt \ 1 \)

continue....

 

[Steven G]

I 2nd induction.

 

[lex]

The problem statement is incorrect.

The infimum should be e (Euler's number). However, if you want to proceed in your way:

The infimum is 2. The supremum is e.

@viswanathan
Perhaps use calculus to show that [MATH]y=\left(1+\tfrac{1}{x}\right)^x[/MATH] is increasing on [MATH] [1,\infty)[/MATH]therefore the minimum value is 2

 

[Deleted member 4993]

The infimum is 2. The supremum is e.

You are absolutely right - I knew that one of these days those Latin words will bite me (I suppose you know that Latin is not my mother tongue!!). Please don't let Jomo see my goof - he is going to send me back to corner for infinitum.

 

[lex]

You are absolutely right - I knew that one of these days those Latin words will bite me (I suppose you know that Latin is not my mother tongue!!). Please don't let Jomo see my goof - he is going to send me back to corner for infinitum.

Very nicely worked!

 

[Eugenio]

You have to use the binomial theorem.

I want to show that the infimum of the set of all numbers of the form [MATH]\bigg(1+\frac{1}{n}\bigg)^n[/MATH] where n=1,2,3,... is 2.

I believe I can prove this if I can show that
[MATH]\bigg(1+\frac{1}{n+1}\bigg)^{n+1} > \bigg(1+\frac{1}{n}\bigg)^n[/MATH]
But I don't know how to show this is true.
Someone please help.

You need to use the binomial theorem. And it's not easy at all. The demonstration is serious business. I found it in "Cálculo diferencial e integral by Ricardo Noriega" 22 years ago I studied from his book and I got 10/10. The demonstration starts on page 25. Good luck!

 

[Steven G]

Subhotosh, back to corner for infinitum