Prove that for $n\ge 2$, the equation $x^n + x-1 = 0$ has a unique root in $[0,1]$. If $x_n$ denotes this root, prove that $(x_n)_n$ is convergent and find its limit.

The limit is $1$. But to find the limit, I need to assume $\lim\limits_{n\to\infty} x_n^n = 0,$ which seems nontrivial (e.g. it doesn't hold for $y_n = 1-1/n$ as $\lim\limits_{n} y_n^n = 1/e,$ even though $0 \le y_n < 1$ for all $n\ge 1$).

The derivative of $f_n(x) = x^n + x-1$ is positive on $[0,1]$, which along with the fact that $f_n(0)f_n(1) < 0$ implies that $f_n(x)$ has a unique root in $[0,1]$.

$x_n$ is convergent because $0 < x_n < 1\Rightarrow 0 < x_{n+1} < 1$ and $x_{n}^{n+1} + x_n - 1 < 0\Rightarrow x_n < x_{n+1}$ as $f_n'(x) > 0$ on $[0,1]$. However, if I do not assume $\lim\limits_n x_n^n = 0$,

I'm not sure how to prove that $\lim\limits_n x_n = 1.$

  • $\begingroup$ You have shown that this sequence is convergent. Therefore, it must converge to a value $x\in[0,1]$. Suppose that $x<1$, and you will get a contradiction, which would mean that $x=1$. $\endgroup$
    – Julián
    Aug 13 at 3:00

5 Answers 5


You have shown:

$x_n\le x_{n+1}$, so the sequence is monotonic increasing and bound above, hence converges.

Next, we need to show: $\lim x_n=1$

$$x_n^n=1-x_n\Rightarrow \lim x_n^n=1-\lim x_n\tag{0}$$

Assume: $\lim x_n=r$ where $0\le r<1$, then Eq.$(0)$ gives: $$\lim x_n^n=1-r>0\tag{1}$$ Since $x_n$ is monotonic increasing, we have $0\le x_n\le r$ $$\Rightarrow 0\le x_n^n\le r^n \Rightarrow 0\le \lim x_n^n\le\lim r^n=0$$ By Squeeze theorem, we have $$\lim x_n^n=0 \tag{2}$$ But this contradicts with Eq.$(1)$, so our assumption is false. Therefore, $$\lim x_n=1$$


For any $n\in\{1,2,3,\ldots\}$ the function $f_n(x)=x^n+x-1$ is increasing and convex on $[0,1]$.
Since $f_n(0)<0$ while $f_n(1)>0$ we have a unique root $x_n\in(0,1)$. By convexity $$ x_n < 1-\frac{f_n(1)}{f_n'(1)}=1-\frac{1}{n+1}\tag{1} $$ and we may notice that $$ f_{n+1}(x_n) = x_n^{n+1}+x_n-1 = x_n(x_n^n+1)-1 = x_n(2-x_n)-1 < -\frac{1}{(n+1)^2}<0\tag{2}$$ such that $x_{n+1}>x_n$ and $\{x_n\}_{n\geq 1}$ is an increasing sequence.
It is bounded by $(1)$, hence $L=\lim_{n\to +\infty} x_n$ exists and it is $\in\left[\frac{1}{2},1\right]$.
Assume that $L<1$. For any $n\geq 1$ we have $x_n^n = 1-x_n$, where $\lim_{n\to +\infty}x_n^n = 0$ while $\lim_{n\to +\infty}(1-x_n)>0$. This is a contradiction, so $L=1$.


I think we could do this by using contradiction, we know $x_n$ converge to $x$, and $0<x\leq 1$, if $0<x<1$, we could see $x_n^n+x_n=1$ is not holded when $n$ is sufficnently large, so we know $x=1$.

  • $\begingroup$ Note that $\lim x_n^n\neq\lim x^n$, so your argument might not hold. $\endgroup$
    – MathFail
    Aug 13 at 4:03
  • $\begingroup$ I mean if $x< 1$ and $\lim x_n \to x$, so if $n$ is sufficiently large $x_n\leq\frac{1+x}{2}<1$,(you can get this by $\epsilon-\delta$),so $x_n^n<(\frac{1+x}{2})^n \to 0$ because $\frac{1+x}{2}<1$, so we can get our conclusion. $\endgroup$ Aug 13 at 13:09

Let $\lim x_n=g.$ For $m\le n$ we have $$x_n^m+x_n-1\ge 0$$ Thus $$g^m+g-1\ge 0,\qquad m\ge 1$$ If $0\le g<1$ then taking the limit $m\to \infty$ gives $g\ge 1.$ Thus $g=1.$

What follows is a complement to the answer by @orangeskid

Let $x_n=1-\delta_n.$ By Bernoulli inequality we get $$0=(1-\delta_n)^n-\delta_n\ge 1-(n+1)\delta_n$$ Hence $$\delta_n\ge {1\over n+1}$$

  • 1
    $\begingroup$ @orangeskid Thanks. What a stupid mistake. I will remove that "solution". $\endgroup$ Aug 13 at 18:57

Write $x_n = \frac{1}{y_n}$, with $y_n > 1$, so $$\frac{1}{y_n^n} + \frac{1}{y_n} = 1$$ or $$1 = y_n^n - y_n^{n-1} = y^{n-1}_n(y_n-1)$$ and with $y_n = 1+\delta_n$ we get $$1 = (1+\delta_n)^{n-1} \cdot \delta_n$$ This implies $\delta_n< 1$. Now, using the Bernoulli inequality $(1+\delta_n)^{n-1} > 1 + (n-1)\delta_n$ we get $$1 >(1+(n-1)\delta_n) \cdot \delta_n = \delta_n + (n-1)\delta_n^2 > n \delta_n^2$$ and so $$0<\delta_n< \frac{1}{\sqrt{n}}$$

$\bf{Added:}$ The inequality $\delta(1+\delta)^{n-1}> n \delta^2$ seems a bit crude, so let's try better. The function $$t \mapsto \frac{t(1+t)^{n-1}}{t^2} = \frac{(1+t)^{n-1}}{t}$$ has on the $[0,1]$ the minimum at $t = \frac{1}{n-2}$, so $$t(1+t)^{n-1}\ge (1+ \frac{1}{n-2})^{n-1}(n-2) \cdot t^2$$ so we get the improved inequality

$$0< \delta_n< \frac{1}{\sqrt{e(n-2)}} $$

In fact we can use the inequalities of the form

$$t(1-t)^{n-1}\ge c_{n,a} t^a$$ to get better estimates for $\delta_n$.

$\bf{Added:}$ Consider the equality

$$f_n(\delta_n) = \delta_n (1+\delta_n)^{n-1} = 1$$

We have $\delta_n > \frac{1}{n}$, since $$f_n(\frac{1}{n}) = \frac{1}{n} ( 1 + \frac{1}{n})^{n-1}< \frac{e}{n} < 1$$ (for $n> 2$). Now, it is not hard to check that

$$f_n(\frac{\log n}{n}) > 1$$ (for $n > 6$) Therefore, we have $$0 < \delta_n < \frac{\log n}{n}$$

a better estimate for $\delta_n$.

$\bf{Added:}$ We have

$$y_n^n - y_n^{n-1} = 1$$ Now, this is the discrete derivative of the function $x \mapsto y^x$, so we get $$y_n^{\xi_n} \log y_n = 1$$ or $$y_n^{\xi_n} \cdot \log y_n^{\xi_n} = \xi_n$$ where $\xi_n \in (n-1, n)$. Now write $$\log y_n^{\xi_n} = z_n$$ and get $$e^{z_n} \cdot z_n = \xi_n$$ This is related to the Lambert function.


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