Proving that the limit of a sequence is $> 0$

Question

Let $u$ be the complex sequence defined as follows :

$u_0=i$ and $ \forall n \in \mathbb N, u_{n+1}=u_n + \frac {n+1-u_n}{|n+1-u_n|} $ .

Consider $w_n$ defined by $\forall n \in \mathbb N,w_n=|u_n-n|$ .

I have to prove that $w_n$ has a limit $> 0$.

Here is what I've proved so far:

• The sequence $Im(u_n)$ is decreasing and bounded by $0$ and $1$ hence convergent with a limit $\in [0;1]$
• $w_n$ is decreasing and bounded by $0$, hence convergent with limit $l \geq 0$

So all I need to prove now is that $l \neq 0$

I tried a proof with contradiction, but could not complete it...

Thanks for you help.

Answer 1

Consider the sequence $z_n = n+1 - u_n$. Then we have $z_0 = 1 - i$, and the recursion

$$z_{n+1} = n+2 - u_{n+1} = 1 + \left(n+1 - u_n - \frac{n+1-u_n}{\lvert n+1-u_n\rvert}\right) = 1 + \left(1 - \frac{1}{\lvert z_n\rvert}\right)z_n.\tag{1}$$

Now it is easy to prove inductively

1. $\lvert z_n\rvert > 1$,
2. $\Re z_n \geqslant 1$,
3. $-1 \leqslant \Im z_n < 0$,
4. $\Re z_{n+1} > \Re z_n$,
5. $\Im z_{n+1} > \Im z_n$.

Writing $r_n = \lvert z_n\rvert$, $\Re z_n = x_n$, $\Im z_n = y_n$, the recursion $(1)$ yields

$$\begin{align} x_{n+1} &= 1 + \frac{r_n-1}{r_n}x_n = x_n + \left(1 - \frac{x_n}{r_n}\right) > x_n \geqslant 1,\\ y_{n+1} &= \frac{r_n - 1}{r_n}y_n, \end{align}$$

since $r_n > 1$ and $y_n \neq 0$. Further we have

$$\begin{align} r_{n+1}^2 &= \left(1 + \frac{r_n-1}{r_n}x_n\right)^2 + \left(\frac{r_n - 1}{r_n}y_n\right)^2\\ &= 1 + 2\frac{r_n-1}{r_n}x_n + \left(\frac{r_n - 1}{r_n}\right)^2(x_n^2+y_n^2)\\ &= 1 + 2\frac{r_n-1}{r_n}x_n + (r_n-1)^2\\ &= r_n^2 - 2(r_n-1)\left(1-\frac{x_n}{r_n}\right)\\ & < r_n^2, \end{align}$$

so the sequence $z_n$ is also bounded. Since the sequences of real resp. imaginary parts are monotonic, the sequence converges, say to $z_\ast = x_\ast + iy_\ast$.

Since the factor by which the imaginary part decreases is bounded away from $1$, we can conclude $y_\ast = 0$.

Since the sequence of real parts is strictly increasing, we have

$$x_\ast > x_1 = \Re \left(1 + \left(1 - \frac{1}{\sqrt{2}}\right)(1-i)\right) = 2 - \frac{1}{\sqrt{2}} > 1.$$

And that means

$$l = \lim_{n\to\infty} \lvert n - u_n\rvert = \lim_{n\to\infty} \lvert z_n - 1\rvert = x_\ast - 1 > 1 - \frac{1}{\sqrt{2}}.$$

Answer 2

Note that $$w_{n+1}=|u_{n+1}-(n+1)|\\ =\left|u_n-n+\frac{n+1-u_n}{|n+1-u_n|}-1\right|\\ =\left|u_n-(n+1)-\frac{u_n-(n+1)}{|u_n-(n+1)|}\right|\\=v_{n}\left|1-\frac{1}{v_{n}}\right|$$ where $$v_{n}=|u_n-(n+1)|$$ Hence $$w_{n+1}=v_n-1$$ because $v_n\ge 0\ \forall \ n\ge 0$. Now, if $u_n=x_n+iy_n,\quad x_n,y_n\in \mathbb{R}\ \forall\ n\ge 0$ Then, $$v_n=\sqrt{(x_n-n-1)^2+(y_n)^2}$$ hence from the recurrence relations, we get $$x_{n+1}=x_n+\frac{n+1-x_n}{v_n}=x_n\left(1-\frac{1}{v_n}\right)+\frac{n+1}{v_n}=x_n\frac{w_{n+1}}{v_n}+\frac{n+1}{v_n}\\ y_{n+1}=y_n\left(1-\frac{1}{v_n}\right)=y_n\frac{w_{n+1}}{v_n}$$

Now, clearly, $v_n\ge 1\Rightarrow \{y_n\} $ is a decreasing sequence. Also, $$x_{n+1}-x_n-1=\frac{n+1-x_n}{\sqrt{(n+1-x_n)^2+y_n^2}}-1\le 0$$ Hence $\{x_{n}-n\}$ is a decreasing sequence. Hence $$v_n=\sqrt{(x_n-n-1)^2+y_n^2}\ge \sqrt{(x_{n+1}-(n+1)-1)^2+y_{n+1}^2}=v_{n+1}$$ i.e. $\{v_n\}$ is a decreasing sequence.

Now, $\frac{w_{n+1}}{v_n}<1\ \forall n \ge 0$ and is a decreasing sequence. Hence let for some $N\in \mathbb{Z^+}$, $\frac{w_{N+1}}{v_N}<\alpha<1$. Then $\frac{w_{n+1}}{v_n}<\alpha$ for all $n\ge N$. Hence, $\forall n\ge N$ $$y_{n+1}<\alpha y_n<\cdots<\alpha^{n+1}\cdot 1$$ Hence, $\lim_{n\rightarrow \infty}y_n\le 0$. But, since $v_n\ge 1,\ y_n\ge 0$. Hence $$\lim_{n\rightarrow \infty}y_n=0$$ Now, $\{n+1-x_n\}$ is increasing $$\Rightarrow \lim_{n\rightarrow \infty}(n+1-x_n)>1+1-x_1=2-\frac{1}{\sqrt{2}}$$ Hence $$\lim_{n\rightarrow \infty}v_n=\lim_{n\rightarrow \infty}\sqrt{(x_n-n-1)^2+(y_n)^2}\\=\sqrt{\lim_{n\rightarrow \infty}(x_n-n-1)^2+\lim_{n\rightarrow \infty}(y_n)^2}=\sqrt{\lim_{n\rightarrow \infty}(-x_n+n+1)^2}=\lim_{n\rightarrow \infty}(-x_n+n+1)>2-\frac{1}{\sqrt{2}}>1$$ Hence $$\lim_{n\rightarrow \infty}w_n=\lim_{n\rightarrow \infty}v_{n-1}-1>0$$

Answer 3

Well the other proofs are quite long... I guess this is shorter.

A simple computation proves that the sequence $n-Re(u_n)$ is increasing.

Now by contradiction, if $l=0$ then $|Re(u_n-n)|=n-Re(u_n)$ converges to $0$

But $n-Re(u_n)$ is increasing.

Hence $\forall n \in \mathbb N, n-Re(u_n) = 0$

That's a contradiction.

I'd like to thank the users who gave a valid solution with a lower bound for $w_n$.