Integer part of $\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$ Find the value of the following infinite series:
$$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$
Now, my doubt is whether it's​ $2$ or it's​ $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it completely attains the value of $3$. 
I would to like to see the proof (formal) as well instead of just intuitions.
 A: Define $x_0:=\sqrt[3]{24}$ and $x_n:=\sqrt[3]{24+x_{n-1}}$ for $n\geqslant 1$. This sequence generates the "infinite series" you have stated. We are interested then whether there is a limit point for $\{x_n\}_{n\in \mathbb{N}}$. First notice that $x_n\leqslant 3$ for all $n\geqslant 0$. This can be proven by induction. For $n=0$ it is obvious. Suppose $x_{n-1}\leqslant 3$ then $$x_n=\sqrt[3]{24+x_{n-1}}\Leftrightarrow x_n^3=24 + x_{n-1}\leqslant 24+3=27\Rightarrow x_n\leqslant 3$$
Hence our sequence is bounded above. Clearly also bounded below by $x_0$. Thus a bounded sequence. Now we show that it is monotone (increasing in this case). 
$$x_n-x_{n-1}=x_n-x_n^3+24=(3-x_n)(x_n^2+3x_n+8)\geqslant 0$$
for all $n$. This follows since $x_n\leqslant 3$ for all $n$ and that the quadratic expression is always positive. Now we have a bounded and monotone sequence so by Bolzano-Weierstrass it must be the case that $\{x_n\}_{n\in\mathbb{N}}$ has a limit point $x$ i.e. $\lim_nx_n=x$. So 
$$\lim_nx_n^3=\lim_n(24+x_{n-1})\Leftrightarrow x^3=24+x\Rightarrow x=3$$
Indeed the infinite series converges to $3$. 
A: $$x=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
$$x=\sqrt[3]{24+x}$$
$$x^3=24+x$$
$$x^3-x-24=0$$
And it's only real root is 3, so:
$$3=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
A: If you assume that the limit exists, it is fairly easy to see that it must be $3$.  However, the existence is not entirely obvious.  To see that the limit exists:
Let $a_n$ be your expression truncated after the $n^{th}$ $24$.  Then the value you want is $\lim_{n\to \infty} a_n$.  
We remark that $a_n=\sqrt[3] {a_{n-1}+24}$.    As $f(x)=\sqrt[3] {x+24}$ is an increasing function we see that $a_n>a_{n-1}$.
We now claim that $a_n<3 \,\,\forall n$.
This is easily confirmed for $n=1$, as $a_1\approx 2.8845$. But if $a_n>3$ then $a_{n-1}=a_n^3-24>3$ so we are done by induction.
Since an increasing bounded sequence of real numbers must approach a limit, we deduce that $\lim_{n\to \infty} x_n$ exists.  Call it $L$.
Now our recursion tells us that $L^3=24+L$ from which it quickly follows that $L=3$.
A: It is equal to $3$.
Let $x=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$. 
Then $x^3-24=x \implies x^3-x-24=(x-3)(x^2+3x+8)=0$. Thus this equation gives $x=3$ as the only real solution.
A: Let $$x=\sqrt[3]{24+\sqrt[3]{24+...}}$$ Substituing $x$ into part of the equation yields
$$x=\sqrt[3]{24+x}$$
Solving for $x$ will give you your answer.
You can do this formally by recursively defining the sequence
$$a_{n+1}=\sqrt[3]{24+a_n}$$
Taking the limit as $n$ goes to infinity yields $$a_{\infty+1}=a_\infty=\sqrt[3]{24+a_\infty}$$
You use the fact that $\infty+1=\infty$ in that last line.
A: Let $(x_n)_{n\in\mathbb{Z}_+}$ the sequence defined by$$x_n=\begin{cases}0&\text{ if }n=0\\\sqrt[3]{24+x_{n-1}}&\text{ if }n\in\mathbb{N}.\end{cases}$$Then your number is $\lim_{n\to\infty}x_n$, if it exists. Now, let $f(x)=\sqrt[3]{24+x}$.
Note that $(\forall n\in\mathbb{Z}_+):x_n<3$. This is clear if $n=0$. Otherwise, if $x_n<3$, then $x_{n+1}<\sqrt[3]{24+3}=3$.
On the other hand, $(x_n)_{n\in\mathbb{Z}_+}$ is an increasing sequence, because, if $0\leqslant x<3$, $f(x)>x$. This can be established observing that $f(x)=x$ if $x=3$ and that $f'(x)=\frac1{3(24+x)^{2/3}}<1$.
Since $(x_n)_{n\in\mathbb{Z}_+}$ is an increasing sequence and it has an upper bound, it converges to some real number $l$. And, since$$(\forall n\in\mathbb{N}):x_n=f(x_{n-1})$$and $f$ is continuous, $l=f(l)$. Now, all that remains to be done is to find the solutions of the euation $f(x)=x$. But$$f(x)=x\iff x+24=x^3,$$and this equation has one and only one real root, which is $3$.
A: Let $$x= \sqrt[3]{24+\sqrt[3]{24+...}}$$ $$x^3 = 24+\sqrt[3]{24+\sqrt[3]{24+...}}$$
Notice $x^3 -24$ will be itself so $$x^3-24=x$$ Solving for $x$ yields $$3, \frac{3+\sqrt{23}i}{2}, \frac{3-\sqrt{23}i}{2}$$ but since we are concerning real solutions, 3 is the answer. It will be infinitely close to 3 amd reaches 3 so the answer is 3. ^^
Edit: That symbol is usually called "floor function". 
A: set 
$$x =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
then
$$x^3 = 24+\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
$$x^3 = 24 + x$$
Shooting for the rational root theorem, consider the divisors of $24$. 2 does not satisfy the equation, clearly as $8 \neq 26$. However, $3^3 = 24 + 3$, so
$$3 =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
Dividing the cubic by $x-3$ yields $x^2 + 3x + 8$ which has no real roots as $9 - 32 = -23 < 0$, so that's our only possible solution.
