Let $x_n=\sqrt[3]{n+1}-\sqrt[3]{n}$. Prove that $(x_n)$ converges. 
Let $x_n=\sqrt[3]{n+1}-\sqrt[3]{n}$. Prove that $(x_n)$ converges.

I managed to prove the sequence converges by using definition of convergence, but initially I thought of using the Monotonic Convergence Theorem to prove it. However, I'm stuck at proving the sequence is decreasing and also bounded. Can anyone lead me in proving these two conditions?
 A: Bounded below is clear, $x_n$ is positive. For decreasing, multiply top and (missing) bottom by $(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}$. We are using the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$. But this is cheating in a way, since as soon as we have done the manipulation, we can compute the limit without using the decreasing bounded below machinery.  
If we are allowed to use the derivative, then it is easy to check that $(x+1)^{1/3}-x^{1/3}$ has negative derivative, and the fact that our sequence is decreasing follows.
A: I would do this a different way, but I will help you with the way you proposed. 
Clearly $\sqrt[3]{n+1}\ge \sqrt[3]{n}$ because $n$ is a positive number, so the sequence is bounded below by zero.
To prove the sequence is decreasing, it suffices to show that $x_{n}\le x_{n-1}$, that is,
$$ \sqrt[3]{n+1} -  \sqrt[3]{n} \le \sqrt[3]{n} - \sqrt[3]{n-1},$$
which is equivalent to
$$ \sqrt[3]{n+1} + \sqrt[3]{n-1}\le 2\sqrt[3]{n}.$$
You might be able to just cube this and show it holds, but the algebra seems hairy. A better way is to note that the cube root function is concave. Then the inequality follows immediately. 
A more efficient way would be to use Taylor series, which allows you to actually compute the limit.
A: $$x_n=\sqrt[3]{n+1}-\sqrt[3]{n}=\frac{(\sqrt[3]{n+1}-\sqrt[3]{n})((n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}})}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}}=\frac{1}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}} \;\;\;\underset{n\to\infty}{\rightarrow}\;\;\;{0}$$
since ${0}<{x_n}<{\dfrac{1}{3n^{\tfrac{2}{3}}}}.$
A: A simple way to prove that $(x_n)$ converges is 
$$x_n=\sqrt[3]{1+n}-\sqrt[3]{n}=\sqrt[3]{n}\left(\sqrt[3]{1+\frac{1}{n}}-1\right),$$
and with the fact that $\sqrt[3]{1+\frac{1}{n}}\sim1+\frac{1}{3n}$ we obtain
$$x_n\sim \frac{1}{3n^{2/3}},$$
and this is sufficient to conclude.
A: You can use a trick to show that this sequence is bounded in the following way. As you cannot pull the third root out of $n+1$ directly, find an expression of the form $\sqrt[3]{n} + f(n)$ such that $(\sqrt[3]{n} + f(n))^3=n+1+g(n)$, such that $g(n)\geq0$ and $f(n)$ goes to zero. Then obviously
\begin{align}
x_n=\sqrt[3]{n+1} - \sqrt[3]{n} 
& \leq \sqrt[3]{n + 1 + g(n)} - \sqrt[3]{n}\\
& = \sqrt[3]{(\sqrt[3]{n} + f(n))^3} - \sqrt[3]{n}\\
& = f(n) \rightarrow 0 \text{ as $n\to\infty$}
\end{align}
Now, to get the $n+1$ part in $n+1+g(n)$ right, choose $f(n)=\frac{1}{3}n^{-2/3}$ and you get 
$$(n^{1/3}+\frac{1}{3}n^{-2/3})^3=n+1+\frac{1}{3n}+\frac{1}{27n^2}\geq n+1.$$
$f(n)$ is clearly bounded and goes to zero as $n\to\infty$, so you're done.
