# Sequence with equidistant terms

Consider the sequence $(u_{n})_{n \in \mathbb{N}}$ given by :

$$u_{0} \in \mathbb{Z} \quad \mathrm{and} \quad \forall n \in \mathbb{N}, \, \vert u_{n+1}-u_{n} \vert = 1.$$

Is the sequence $(w_{n})_{n \in \mathbb{N}} = \displaystyle \Big( \frac{u_{n}}{n+1} \Big)_{n \in \mathbb{N}}$ convergent ?

I have proved that $(w_{n})_{n \in \mathbb{N}}$ is bounded ($\forall n \geq n_{0}, \, 0 \leq w_{n} \leq 2$). I tried to see whether it is increasing or decreasing but we do not have enough information on $(u_{n})_{n \in \mathbb{N}}$ to conclude. I also tried to prove that $(w_{n})_{n \in \mathbb{N}}$ is a Cauchy sequence but I don't think it is one. My intuition is that $(w_{n})_{n \in \mathbb{N}}$ is convergent but I can't prove it.

Edit : To prove that $(w_{n})_{n \in \mathbb{N}}$ is bounded, here is what I did : let $n \in \mathbb{N}$,

$$u_{n} - u_{0} = \sum_{k=1}^{n} u_{k}-u_{k-1}$$

Then,

\begin{align*} \vert u_{n}-u_{0} \vert &= {} \Big\vert \sum_{k=1}^{n} u_{k}-u_{k-1} \Big\vert \\[2mm] &\leq \sum_{k=1}^{n} \vert u_{k}-u_{k-1} \vert = n \\ \end{align*}

As a consequence, $\vert u_{n}-u_{0}\vert \leq n \leq n+1$. It leads to : $\vert u_{n} \vert \leq \vert u_{0} \vert + (n+1)$. Therefore :

$$\bigg\vert \frac{u_{n}}{n+1} \bigg\vert \leq \frac{\vert u_{0} \vert}{n+1} + 1$$

Since the sequence $\displaystyle \Big( \frac{\vert u_{0} \vert}{n+1} \Big)_{n \in \mathbb{N}}$ is convergent to $0$, there exist a $n_{0} \in \mathbb{N}$ such that :

$$\forall n \geq n_{0}, \, \frac{\vert u_{n} \vert}{n+1} \leq 2.$$

Edit 2 : To prove that $(w_{n})_{n \in \mathbb{N}}$ is a Cauchy sequence, here is what I tried : let $(p,n) \in \mathbb{N}^{2}$ such that $p > n$,

\begin{align*} \frac{u_{p}}{p+1} - \frac{u_{n}}{n+1} &= {} \frac{(n+1)u_{p} - (p+1)u_{n}}{(n+1)(p+1)} \\[2mm] &= \frac{(n+1)\bigg( \displaystyle \sum_{k=n+1}^{p} (u_{k}-u_{k-1}) + u_{n} \bigg) - (p+1)u_{n}}{(n+1)(p+1)} \\[2mm] &= \frac{\displaystyle (n+1)\sum_{k=n+1}^{p} (u_{k}-u_{k-1}) - (p-n)u_{n}}{(n+1)(p+1)} \\ \end{align*}

Therefore,

$$\Bigg\vert \frac{u_p}{p+1} - \frac{u_n}{n+1} \Bigg\vert \leq \frac{(p-n)(n+1+\vert u_{n} \vert)}{(n+1)(p+1)}$$

But I couldn't go any further.

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Well if it wasnt cauchy it couldnt possible converge –  Kamster Aug 26 at 7:57
Could you write your proof for boundedness please? –  Surb Aug 26 at 7:59
Also sequence isnt necessarly increasing or decreasing it could just up one and down one for whole sequence –  Kamster Aug 26 at 8:02
@Surb : I edited my post. –  Odile Aug 26 at 8:16
My intuition says that either $u_n=o(n)$ or $u_n\approx \pm n$ (with a very imprecise notation). In the secondo case I wonder if $\{w_n\}$ may oscillate... –  Siminore Aug 26 at 8:33

Your sequence does not converge in general. To see this, first note that the requirements on $\left(u_{n}\right)_{n\in\mathbb{N}_{0}}$ can equivalently be formulated as $$\text{For all }n\in\mathbb{N}_{0}\text{, there is }\varepsilon_{n}\in\left\{ \pm1\right\} \text{ such that }u_{n+1}=u_{n}+\varepsilon_{n}.$$ Now, we choose $u_{n}:=0$ and choose the $\left(\varepsilon_{n}\right)_{n\in\mathbb{N}_{0}}$ as follows:

1. Choose $\varepsilon_{0}:=1$.
2. If $\varepsilon_{j}$ has already been chosen for $j=0,\dots,\sum_{j=0}^{m-1}\left(2^{j}+2^{j+1}\right)$ for some $m\in\mathbb{N}_{0}$, choose \begin{eqnarray*} \varepsilon_{j} & := & -1\text{ for }j=\sum_{j=0}^{m-1}\left(2^{j}+2^{j+1}\right)+1,\dots,\sum_{j=0}^{m-1}\left(2^{j}+2^{j+1}\right)+2^{m},\\ \varepsilon_{j} & := & 1\text{ for }j=\sum_{j=0}^{m-1}\left(2^{j}+2^{j+1}\right)+2^{m}+1,\dots,\sum_{j=0}^{m}\left(2^{j}+2^{j+1}\right). \end{eqnarray*}

This basically means the following (you will see the pattern after finitely many terms): \begin{eqnarray*} u_{0} & = & 0,\\ u_{1} & = & 1,\\ u_{2} & = & 0,\\ u_{3} & = & 1,\\ u_{4} & = & 2=2^{1},\\ u_{5} & = & 1,\\ u_{6} & = & 0,\\ u_{7} & = & 1\\ u_{8} & = & 2\\ u_{9} & = & 3,\\ u_{10} & = & 4=2^{2},\\ \vdots \end{eqnarray*} Inductively, you can show that \begin{eqnarray*} u_{1+\sum_{j=1}^{m-1}\left(2^{j}+2^{j+1}\right)+2^{m}} & = & 0,\\ u_{1+\sum_{j=1}^{m}\left(2^{j}+2^{j+1}\right)} & = & 2^{m+1}. \end{eqnarray*} The first equation tells us that $0$ is an accumulation point of $w_{n}$, so if it converges, it converges to $0$.

But: $$w_{1+\sum_{j=1}^{m-1}\left(2^{j}+2^{j+1}\right)}=\frac{2^{m}}{2+\sum_{j=1}^{m-1}\left(2^{j}+2^{j+1}\right)}=\frac{2^{m}}{3\cdot2^{m}-4}\xrightarrow[m\rightarrow\infty]{}\frac{1}{3}\neq0.$$ Hence, the sequence is not convergent.

EDIT: The intuition here is that the exponential length of the "intervals"

$$\sum_{j=0}^{m-1}\left(2^{j}+2^{j+1}\right),\dots,\sum_{j=0}^{m}\left(2^{j}+2^{j+1}\right)$$

allows you to "forget" the "history" up to that point.

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I thought about this "pyramidal" sequence but after some time, I also thought that the associated sequence $(w_{n})_{n}$ would converge to $0$. I didn't think about it enough. Thanks for your answer ! –  Odile Aug 26 at 9:28