Show that if n belongs to N, and: An: = (1 + 1/n)^n then An > An+1 for all natural n

Sep 2010
32
0
Show that if n belongs to N, and:

An: = (1 + 1/n)^n

then An < An+1 for all natural n. (Hint, look at the ratios An+1/An, and use Bernoulli's inequality)

I think i have a vague idea of what to do here, like im sure induction is involved in this proof/ However, im unsure how bernoullis inequality and the ratios help in the proof. Can anyone help me please?
 
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Sep 2010
32
0
made a mistake, that i fixed, it's supposed to be an < an+1 as opposed to the other way around.
 
Sep 2010
32
0
I started the problem by doing the ratios like it suggests and ended up with:

[(1+1/(n+1))((1+1/(n+1)^n)]/(1+1/n^n) > 1

Where do i go from here, and when do i use bernoullis inequality.
 

Plato

MHF Helper
Aug 2006
22,506
8,663
Re: Show that if n belongs to N, and: An: = (1 + 1/n)^n then An > An+1 for all natur

Show that if n belongs to N, and:
An: = (1 + 1/n)^n
then An < An+1 for all natural n. (Hint, look at the ratios An+1/An, and use Bernoulli's inequality)?
Suppose that $n\ge 2$ then
$ \begin{align*}\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^n &=\left(1-\dfrac{1}{n^2}\right)^n \\\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^n&\ge1+n\left(\dfrac{-1}{n^2}\right)= \left(1-\dfrac{1}{n}\right)\text{ Bernoulli's inequality}\\ \left(1+\dfrac{1}{n}\right)^n&\ge\left(1-\dfrac{1}{n}\right)^{1-n}\\&\ge\left(\dfrac{n}{n-1}\right)^{n-1}=\left(1+\dfrac{1}{n-1}\right)^{n-1} \end{align*}$


I know this post is old but unanswered.
 
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