Show that an expression approaches an integer

[John Greger]

TL;DR

Is this limit an integer?

I came across a rather strange thing in an introductory class I still don't understand.

There was a statement that $$lim_n (2+ \sqrt(2))^n $$ is an integer. I recalled that I never understood this and just recently tried to take the limit but just get that the expression diverge? Which I think makes sense, so how can this expression be an integer?

 

[fresh_42]

Summary:: Is this limit an integer?

... so how can this expression be an integer?

$n \to 0$ or $n\to -\infty$

 

[John Greger]

$n \to 0$ or $n\to -\infty$

Sorry about that! Will try to edit the post, should be as n goes to infinity!

 

[John Greger]

I could do a Binomial expansion and argue that every ${n \choose k}$ are integers so we reduce the problem to arguing that the sum of 2 to some power of n should an integer but it does not bring me so much closer...

 

[PeroK]

I could do a Binomial expansion and argue that every ${n \choose k}$ are integers so we reduce the problem to arguing that the sum of 2 to some power of n should an integer but it does not bring me so much closer...

Why would the limit be finite?

 

[Mark44]

Why would the limit be finite?

Indeed.
In fact, for any real number $\epsilon \ge 0$, $\lim_{n \to \infty} (2 + \epsilon)^n = \infty$. This should be easy to prove.

 

[Infrared]

Perhaps the intended question why the difference between $(2+\sqrt{2})^n$ and the nearest integer goes to zero when $n\to\infty.$ This is because $(2+\sqrt{2})^n+(2-\sqrt{2})^n$ is always an integer, and the latter term goes to zero.

 

[John Greger]

Perhaps the intended question why the difference between $(2+\sqrt{2})^n$ and the nearest integer goes to zero when $n\to\infty.$ This is because $(2+\sqrt{2})^n+(2-\sqrt{2})^n$ is always an integer, and the latter term goes to zero.

This makes sense!