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Given that $x_1 = 1$ and $x_{n+1}=x_{n}+ 1/x_{n}^{2}$. Find the limit of the sequence.

Let $ c $ be the limit of the sequence, then $ c=c+\frac{1}{c^2} $, that means $ \frac{1}{c^2}=0 $. That can't be like that.What is wrong???

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  • $\begingroup$ What is the initial value $x_1$? What happens if $x_1 = -1$? $\endgroup$ – user99914 Nov 9 '13 at 19:57
  • $\begingroup$ If you write $x_{n+1} = (x_n^3 + 1)/(x_n^2)$ what will happen as n $\rightarrow \infty$? John shows you one case. What happens if $x_1$ = 1/2? -1/2? 1? 2? You've got several cases here. $\endgroup$ – Betty Mock Nov 9 '13 at 20:05
  • $\begingroup$ The initial value is $x_{1}=1$ $\endgroup$ – evinda Nov 9 '13 at 20:08
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    $\begingroup$ Have you considered that the sequence might not converge to a real number? $\endgroup$ – Daniel Fischer Nov 9 '13 at 20:19
  • $\begingroup$ I don't know..but the exercise asks me to find the limit,so I suppose that the sequence converges to a real number. $\endgroup$ – evinda Nov 9 '13 at 20:28
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Suppose the sequence converges to $c$. Can you see how your definition of the sequence implies that $c>0$?

Then for any $\epsilon>0$, the sequence eventually exceeds $c-\epsilon$.

If $c-\epsilon < x_n \le c$ then $x_n+\dfrac 1 {x_n^2}>{?}+\dfrac 1 {?}$. Can you fill in the blanks and see how this leads to a contradiction?

Showing the sequence increases without bound

Every increasing sequence of real numbers that is bounded above converges. As $(x_n)$ is an increasing sequence of real numbers that does not converge, it must be unbounded above.

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  • $\begingroup$ If $ c-ε<x_{n}<=c $ then $ x_{n}+\frac{1}{x_{n}^2}>c-ε+c^2$ . What does this mean? That the sequence isn't bounded above? $\endgroup$ – evinda Nov 9 '13 at 21:52
  • $\begingroup$ @evinda, you got the last term wrong. Think about what you can set $\epsilon$ to. $\endgroup$ – dfeuer Nov 9 '13 at 22:02
  • $\begingroup$ What do you mean?? :/ $\endgroup$ – evinda Nov 10 '13 at 0:11
  • $\begingroup$ You made a minor error. The $c^2$ term should be $1/c^2$. Once you've fixed that, see if you can choose an $\epsilon$ for which the inequality does not hold, which will give you a contradiction. $\endgroup$ – dfeuer Nov 10 '13 at 0:32
  • $\begingroup$ Oh, sorry..you're right! Could you give a hint to choose an ε..I got really stuck right now. :/ $\endgroup$ – evinda Nov 10 '13 at 0:49
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Notice $$x_{n+1}^3 = x_n^3 (1 + \frac{1}{x_n^3})^3 = x_n^3 + 3 + \frac{3}{x_n^3} + \frac{1}{x_n^6}$$ We have $x_{n+1}^3 - x_n^3 > 3$ for all $n$ and hence we can bound $x_n$ from below

$$x_n^3 \ge x_1^3 + 3(n-1) = 3n-2 \quad\implies\quad x_n \ge \sqrt[3]{3n-2}$$

As a result, the sequence $x_n$ diverges.

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  • $\begingroup$ Minor nitpick: you should have $x_n^2$ in the denominator $\endgroup$ – Empy2 Nov 10 '13 at 7:24
  • $\begingroup$ @Michael $x_n + \frac{1}{x_n^2} = x_n (1+ \frac{1}{x_n^3})$ $\endgroup$ – achille hui Nov 10 '13 at 7:42
  • $\begingroup$ @achillehui Your inequality is "equivalent" to assume a 'slow' variation of $\displaystyle{\large x_{n}}$ and solve $\displaystyle{\large{{\rm d}x_{n} \over \dd n} = {1 \over x_{n}^{2}}}$ with the condition $\displaystyle{\large x_{1} = 1}$. That yields $\displaystyle{\large x_{n} \sim \sqrt[3]{3n - 2}}$. $\endgroup$ – Felix Marin Nov 11 '13 at 1:25
  • $\begingroup$ @FelixMartin Yup, that's actually the reason why I look at the behavior of $x_n^3$ instead of $x_n$. $\endgroup$ – achille hui Nov 11 '13 at 4:29
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Since $x_{n+1}> x_n$ here, it implies that the limit of $x_n$ is either a finite number or goes to infinity.

The first case is impossible, as $$\lim_{n\rightarrow\infty} x_n =l \in (1, \infty) \Rightarrow l+\frac{1}{l^2}=l \Rightarrow 1=0\text{(contradiction)}$$ Thus, $\lim_{n\rightarrow\infty} x_n = \infty$.


Added:(Proving that the sequence isn't bounded above)Since the sequence $x_n$ is increasing, then if it is convergent, it should have upper boundary and its limit is equal to its least upper bound (This means that if $l$ is an upper bound for $x_n$, for any other upper bound $L$ of $x_n$, we have $l \le L$). Suppose the limit of $x_n$ is $l$, which implies $x_n \lt l$. Also,by the definiton of convergent sequences, we have: $$\forall \epsilon\gt0, \exists N\gt0,\text{such that, }n\gt N, \left|x_n-l\right|\lt \epsilon$$

hence $l-\epsilon\lt x_n\lt l$. Let $\epsilon = \frac{1}{l^2}$,then by the given formula above: $$x_{n+1} = x_n + \frac{1}{x_n^2} \gt l-\epsilon + \frac{1}{l^2} = 1-\frac{1}{l^2}+\frac{1}{l^2} = l$$

which contradict to the fact that $x_{n+1} \lt l$. Hence, the upper bound do not exist,$x_n$ is divergent.

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  • $\begingroup$ Could I also show that the limit of $x_{n}$ goes to infinity by proving that the sequence isn't bounded above? I tried to prove this doing the following: Since $x_{1}=1$ and the sequence is incresing, that means that $x>=1$ for each n. So $x_{n+1}=x_{n}+\frac{1}{x_{n}^2}>=1+1=2>=1$. Are my thoughts right? $\endgroup$ – evinda Nov 10 '13 at 0:32
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    $\begingroup$ @evinda, the easiest thing is to use the fact that the sequence is increasing but does not have a limit. There are various ways to prove that it has no limit. sundaycat's approach uses continuity of $x\mapsto x+\frac 1{x^2}$ on $[1,\infty)$. My approach is a direct $\epsilon$-$N$ proof. You can do it in various ways. $\endgroup$ – dfeuer Nov 10 '13 at 0:37
  • $\begingroup$ @evinda In order to prove that the sequence isn't bounded you need to create an contradiction here, I just edited my answer. check the Added part.(Note, my second solution is the same as dfeuer's) $\endgroup$ – SundayCat Nov 10 '13 at 6:03
  • $\begingroup$ @sundaycat, you don't need to reproduce my approach to prove the sequence is unbounded. Your original proof implicitly uses continuity to show that the sequence does not converge: if $(x_n)$ converges to $c$, then $(x_n)$ must converge to $c+\frac1{c^c}$. Once you've proven the sequence does not converge, the fact that it is increasing immediately proves that it increases without bound. $\endgroup$ – dfeuer Nov 10 '13 at 6:11
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\large\tt Hint:$ $$ x_{n} = x_{n - 1} + {1 \over x_{n - 1}^{2}} = x_{n - 2} + {1 \over x_{n - 2}^{2}} + {1 \over x_{n - 1}^{2}} = \cdots = x_{1} + {1 \over x_{1}^{2}} + {1 \over x_{2}^{2}} + \cdots + {1 \over x_{n - 1}^{2}} $$

$$ x_{n} = x_{1} + \sum_{k = 1}^{n - 1}{1 \over x_{k}^{2}}\,\qquad n \geq 2 $$ If $\lim_{n \to \infty}x_{n}$ is finite, the 'generic serie term' should go to zero when $n \to \infty$: $\quad\lim_{k \to \infty}\pars{1/x_{k}^{2}} = 0$ which contradicts the initial statement. Then, $\lim_{n \to \infty}x_{n} = \infty$.

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