Show that $(x+ \sqrt{x})^n$ is arbitrarily close to an integer for $x \leq 3$ as $n$ approaches $\infty$ So basically I want to show two things, firstly that $$(2+\sqrt{2})^n$$ approaches an integer as $n$ gets large. And secondly that this is only true for $$\lim_{n \to \infty }(x+\sqrt{x})^n, x \leq 3.$$
To show the first one, I tried to expand the expression using the binomial theorem. Arguing that the binomial coefficients are positive integers so it remains to show that a sum of $2$'s to some power of $n$ is an integer there I'm pretty much stuck...
 A: Hint: what is $(x+\sqrt{x})^n +(x-\sqrt{x})^n$? How big can $x$ get to ensure $(x-\sqrt{x})^n$ decays as $n\to\infty$?
A: This is called a Pisot number, consider $r=2+\sqrt{2}$ and its conjugate $\bar r=2-\sqrt{2}$
Notice that $|\bar r|<1$ so that $(\bar r)^n\to 0$.
Now let's have a polynomial with $r$ and $\bar r$ as roots $P(t)=(t-r)(t-\bar r)=t^2-4t+2$
This is also the characteristic equation of a linear recurrence relation $$u_{n+2}=4u_{n+1}-2u_n$$
Notice that if $u_0$ and $u_1$ are integers, then by induction $u_n$ is an integer for all $n$.
The general solutions are expressed as $$\alpha r^n+\beta \bar r^n$$
Initial terms are $\begin{cases}u_0=\alpha+\beta\\u_1=2(\alpha+\beta)+\sqrt{2}(\alpha-\beta)\end{cases}$
So let's get rid of the square root (need to be an integer), for instance with $\alpha=\beta=1$.
We have proved that $r^n+\bar r^n\in\mathbb Z$ and since $\bar r^n\to 0$ then $r^n$ gets close to an integer as $n$ goes to infinity.

You can do it with any $r=x+\sqrt{x}$ as long as :

*

*$|x-\sqrt{x}|<1$, which implies $x<\frac{3+\sqrt{5}}{2}$ (so $x=3$ do not work)


*we also need $P(t)$ to be a polynomial with integer coefficients $\begin{cases}2x\in\mathbb Z\\x^2-x\in\mathbb Z\end{cases}$ in order to make induction for $u_n\in\mathbb Z$ work.
But this condition pretty much restricts $x$ to $0,1,2$

Let $f(x)=x-\sqrt{x}$ then $f'(x)=1-\frac 1{2\sqrt{x}}$
So $f$ has a minimum in $x=\frac 14$ (i.e. $f'(x)=0$) whose value is $f(\frac 14)=-\frac 12$, which solve the case $f(x)>-1$.
Now $f(x)<1\iff (x-1)<\sqrt{x}$ but since $x\ge 0$ (domain of $f$) we don't loose equivalence squaring this inequality.
Thus $(x-1)^2<x\iff x^2-3x+1<0$ solving the quadratic leads to $x$ between the roots $\frac{3\pm\sqrt{5}}{2}$, but we are not interested in the lower one since it is $>\frac 14$ so $f\nearrow$ there.
$$|x-\sqrt{x}|<1\iff 0\le x<\frac{3+\sqrt{5}}{2}$$
