Show that $u_{n+1}-u_{n}$ converge to $0$ with $u_n$ a real bounded sequence Could you please help me to solve this problem :

Let $u_n$ a real bounded sequence, $v_n=u_{n+1}-u_{n}$ and $w_n=v_{n+1}-v_n$.
Assume $(w_n)_{n\in \mathbb{N}}$ converge. Show that $(w_n)_{n\in \mathbb{N}}$ tends to $0$ and $(v_n)_{n\in \mathbb{N}}$ too.

I tried severals methods here,

*

*First one (credulous),

$(w_n)_{n\in \mathbb{N}}$ converge: Denote $l$ the limit,  so there exist \epsilon >0 such that for $n \ge N \implies \vert w_n -l \vert < \epsilon$ and use inequalities.

*

*Cauchy criterion

Credulous too because this criterion doesn't not give us the limit.
So, in fact it seems that classics methods doesn't work (like sequence extracted) but perhaps I am wrong.
Thank you in advance,
NB. Perhaps this result can be useful here :
If a sequence $(A_n)$ is real, and $A_{n+1}-A{_n} \rightarrow 0$ as $n\rightarrow +\infty$
then all accumulation points ​​of the sequence $(A_n)$ is an interval.
 A: Your first method looks like it will work fine.
The easiest way to do such proofs is to forget the fine details for a minute, and think about an actual example. If $w$ converges to some limit, the gaps between $v$ are close to constant, so $u$ keeps increasing or decreasing by a constant rate, and is therefore unbounded.
Now prove it:
Suppose $w_n$ converges to $l$, and assuming for now $l>0$ ($l<0$ will work similarly).
Choose some $0<\epsilon<l$. You know for some $N$ that $v_{n+1}-v_n>l-\epsilon$ for all $n\geq N$.
Summing the LHS for a range of $n$ values gives $v_{N+k} - v_N > k(l-\epsilon)$ for all $k$, ie $v_{N+k} > k(l-\epsilon)+v_N = C > 0$ for all $k$.
So for $j>=N+k$, $u_{j+1}-u_j>C$.
This means $u$ is unbounded, which is a contradiction.
Eliminating $l<0$ in the same way gives $l=0$.
A: Suppose that $$L = \lim_{n \to \infty} w_n > 0$$
Then there exists an integer $N$ such that whenever $n \ge N$, we have
$$u_{n + 2} - u_n = v_{n + 1} - v_n = w_n > \frac L 2$$
In particular, after a short induction, this implies that for all $k$, we have
$$u_{N + 2k} > \frac L 2 k - u_N$$
(each time we move $2$ indices farther up, we increase by $L/2$; this happens $k$ times). Since $u_N$ is fixed, this implies that $u_{n}$ is unbounded, giving the first result.

Here is a sketch of a proof of the second proposition: Note that we have the following relations:
\begin{align*}
w_{n + 1} - w_n &= v_{n + 2} - v_{n + 1} + v_{n + 1} - v_n = v_{n + 2} - v_n\\
w_{n + 2} - w_{n + 1} + w_n &= v_{n + 3} -v_n \\
&\vdots \\
w_{n + k} - w_{n + k - 1} + \dots \pm w_n &= v_{n + k + 1} - v_n
\end{align*}
after telescoping appropriately. If $n$ is large enough, the left hand side can be bounded by a small $\epsilon$, since we're considering an alternating sum of very small numbers (using the convergence of $w_n$ to $0$). Thus the right hand side can be bounded, and so the sequence $(v_n)$ is Cauchy. A similar argument to the first part then shows that the limit is zero.
