If the distance between two consecutive terms in a sequence with bounded partial sum goes to 0, can we say the sequence converges to 0? This is a problem that I met in a recent exam:
If $(x_n)$ is a real-valued sequence and $(x_n)$ is bounded. Suppose that
$$x_n-2x_{n+1}+x_{n+2}\to 0 \ \text{as }n\to \infty.$$ Show that $\lim_{n\to \infty} (x_n -x_{n+1})= 0$.  
Here is what I got during the exam. Let $a_n = x_n - x_{n+1}$ and $S_n = \sum_{k=1}^n a_n$, so the problem becomes: If $(S_n)$ is bounded and $\lim_{n\to \infty} (a_n-a_{n+1}) = 0$, then $a_n\to 0$. The boundedness of partial sum plays an important role here. It is well-known that Cauchy-criterion can't be satisfied if we only assume $\lim_{n\to \infty} (a_n-a_{n+1}) = 0$, we need to gurantee this distance decrease to $0$ at a fast rate with order such as $\frac{1}{n^{1+\varepsilon}}$ or $r^n$ with $|r|<1$. My idea is to prove this result by contradiction. Suppose there exists an $\varepsilon>0$, for all $N\in \mathbb{N}$, there exists a number $n_1$ such that
$$|a_{n_1}|\ge \varepsilon.$$ So we can inductively choose a subsequence of $(a_n)$ with absolute sum blowing up, however it doesn't help since $(a_n)$ can be negative. Then I figure there might be some counter-example such that $(a_n)$ is oscillating between $(-\alpha, +\alpha)\supset(-\varepsilon, \varepsilon)$ and $a_n-a_{n+1}$ is of order $1/n$, so the cancellation among $(a_n)$ can still ensure a bounded partial sum. But I failed in either way. Can anyone provide me with some help on this problem? Thank you.
 A: Here is a weird proof. 
We are told that $|x_n|\leq c$ for some constant $c$. 
Suppose there is an $\epsilon>0$ such that $|x_n-x_{n+1}| \geq \epsilon$ for infinitely many $n$ (we reach a contradiction). Let $\{n[k]\}_{k=1}^{\infty}$ be a subsequence of indices for which $|x_{n[k]}-x_{n[k]+1}|\geq \epsilon$ for all $k \in \{1, 2, 3,…\}$.  Fix $r\geq 10$ as a positive integer such that $(r-1)\epsilon >2c$ and consider the $r$-dimensional sequence $$\{(x_{n[k]}, x_{n[k]+1}, …, x_{n[k]+r-1})\}_{k=1}^{\infty}$$ This is bounded in $\mathbb{R}^r$ so the Bolzano-Wierstrass Theorem says there is a convergent subsequence defined by indices $n[k_m]$ such that 
$$ (x_{n[k_m]}, x_{n[k_m]+1}, …, x_{n[k_m]+r-1})\rightarrow (y_1, y_2, …, y_{r})$$
for some $(y_1, ..., y_r) \in \mathbb{R}^r$. Also, we must have $|y_1-y_r|\leq 2c$. 
On the other hand we know $x_n-2x_{n+1}+x_{n+2}\rightarrow 0$ so we must have 
$$ y_i - 2y_{i+1} + y_{i+2} = 0 \quad \forall i \in \{1, …, r-2\}$$
It follows that $y_i$ has the form: 
$$ \boxed{y_i = A + Bi \quad \forall i \in \{1, 2, 3, ..., r\}}$$
for some constants $A, B$. 
On the other hand we have $$|y_1-y_2|\geq \epsilon$$ and so 
$$ |\underbrace{(A+B)}_{y_1} - \underbrace{(A+2B)}_{y_2}|\geq \epsilon \implies |B|\geq \epsilon $$
But that means 
$$ |y_1-y_r| = |\underbrace{(A+B)}_{y_1}-\underbrace{(A+rB)}_{y_r}| =(r-1)|B|\geq (r-1)\epsilon >2c $$ 
which contradicts $|y_1-y_r|\leq 2c$.  $\Box$
A: I have a proof that I think is similar to Michael's.
Given that the sequence is bounded, we know that $\exists D \in \mathbb{R}, \forall m, n \in \mathbb{N},  \left| x_n - x_m \right| < D$.
Given that $x_{m+2} - 2x_{m+1} + x_m \to 0$, we know that $\forall \delta >0, \exists N\in \mathbb{N}, \forall m > N, \left| x_{m+2} - 2x_{m+1} + x_m  \right| < \delta$.
If for all such $\delta$ and $N$ we also have that $\forall m > N, \left| x_{m+1} - x_m  \right| < \delta$, then we're done.
If not, 
For some $n> N$, let $\left| x_{n+1} - x_n \right| = \epsilon \geq \delta$ and define $k = \lfloor \frac{\epsilon}{\delta} \rfloor$.
We have that $\left| x_{n+1} - x_n \right| \geq k\delta$.
Furthermore, we know that $\left| \left(x_{n+2} - x_{n+1}\right) - \left(x_{n+1} - x_{n}\right)\right| < \delta$, so we have that $\left| x_{n+2} - x_{n+1} \right| > (k-1)\delta$.
Continuing to use that inequality, for  $0 < j \leq k$, we have $\left| x_{n+1+j} - x_{n+j} \right| > (k-j)\delta$
We also have that $0 \leq j \leq k \rightarrow \operatorname{sign}\left(x_{n+j+1} - x_{n+j}\right) = \operatorname{sign}\left(x_{n+1} - x_{n}\right)$.
Therefore $D > \left| \left(x_{n+k+1} - x_{n}\right)\right| > \sum_{i=0}^k i \delta = \frac{k(k+1)}{2} \delta \geq \frac{k}{2} \frac{\epsilon}{\delta} \delta = \frac{k}{2} \epsilon \geq \frac{(\frac{\epsilon}{\delta})-1}{2} \epsilon$.
Hence, we have $\epsilon^2 -\delta \epsilon < 2 \delta D$.
So $\forall n > N, \left( \epsilon - \frac{\delta}{2}\right)^2 = \epsilon^2 -\delta \epsilon +\frac{\delta^2}{4} < 2 \delta D +\frac{\delta^2}{4}$
Which means that $\forall n > N, \epsilon < \frac{\delta}{2} + \sqrt{2 \delta D +\frac{\delta^2}{4}} < \delta + \sqrt{2 \delta D} $
Therefore, by choosing $\delta$ small enough, we can force $\left| x_{n+1} - x_n \right|$ to be as small as we like.
In particular, $\forall \epsilon > 0$, choose $\delta = \min(2D, \frac{\epsilon^2}{8D}) > 0$ (if $D=0$, then the sequence is constant and convergence is obvious), so that $\delta < \sqrt{2 \delta D}$.
Then $\exists N\in \mathbb{N}, \forall m > N, \left| x_{m+2} - 2x_{m+1} + x_m  \right| < \delta$, and the derivation above shows that $\forall m > N, \left| x_{n+1} - x_n \right| <  \delta + \sqrt{2 \delta D} \leq 2\sqrt{2 \delta D} \leq \sqrt{8 \delta D} \leq \epsilon$
