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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?

Zev Chonoles
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Idonknow
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5 Answers5

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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.

André Nicolas
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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.

Potato
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  • I actually found a nice way around that “hairy algebra”. by AM-GM, we have $3\cdot \sqrt[3]{abc} \leq a+b+c$ for positive reals a, b, c. Cubing the left hand side then gives $(n+1) + 3\sqrt[n] {(n+1)(n+1)(n-1)} + 3\sqrt[n]{(n+1)(n-1)(n-1)} + (n-1) \leq (n+1) + (3n + 1) + (3n - 1) + (n-1) = 8n$ – Lukas Juhrich Dec 07 '20 at 22:51
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$$x_n=\sqrt[3]{n+1}-\sqrt[3]{n}=\frac{(\sqrt[3]{n+1}-\sqrt[3]{n})((n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}})}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}}=\frac{1}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}} \;\;\;\underset{n\to\infty}{\rightarrow}\;\;\;{0}$$ since ${0}<{x_n}<{\dfrac{1}{3n^{\tfrac{2}{3}}}}.$

M. Strochyk
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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.

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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.

Elmar Zander
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  • I'm not quite understand how you conclude the sequence is $(x_n)$ is bounded . – Idonknow Feb 25 '13 at 14:17
  • That $(x_n)$ is bounded from below, is trivial, as $x_n$ is clearly positive for all $n$. Then, for $f(n)=\frac13 n^-{2/3}$, you have $x_n\leq f(n)$ and $f(n)\leq\frac13$ for all $n$. Thus, $(x_n)$ is also bounded from above. – Elmar Zander Feb 25 '13 at 15:50
  • I expanded the answer also a little bit. I hope it's clearer now. BTW: To get the expression for $f(n)$ just consider $(n^{1/3}+f(n))^3=n+3f(n)n^{2/3}+3f(n)^2n^{1/3}+f(n)^3$ and solve $3f(n)n^{2/3}=1$ in order to fit the requirements for $f$ and $g$. – Elmar Zander Feb 25 '13 at 16:01
  • ooh, I understand now. – Idonknow Feb 25 '13 at 17:40