# Given that for each $n,\;x_n^n + x_n-1= 0,$ is $(x_n)_n$ convergent?

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 $$\lim\limits_n x_n = 1.$$

• 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 $x=1$. Aug 13 at 3:00

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 $$0, if $$0, we could see $$x_n^n+x_n=1$$ is not holded when $$n$$ is sufficnently large, so we know $$x=1$$.

• Note that $\lim x_n^n\neq\lim x^n$, so your argument might not hold. Aug 13 at 4:03
• 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. 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 $$\delta_n\ge {1\over n+1}$$

• @orangeskid Thanks. What a stupid mistake. I will remove that "solution". 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.