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.
 A: 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:


*

*Choose $\varepsilon_{0}:=1$.

*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.
