4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ $\lim_{n\to \infty} a_n = \lim_{n-1\to \infty} a_n = \lim_{n\to \infty} a_{n+1}$. $\endgroup$ Commented Sep 24, 2019 at 22:36
  • $\begingroup$ @fleablood I agree with you. If the limit of $(a_n)$ exists, then it must be $0$. But how do we show the limit exists? $\endgroup$ Commented Sep 24, 2019 at 22:45
  • $\begingroup$ A different way to interpret this question is that if the 2nd order forward differences tend to $0$, then the 1st order forward difference will as well (as long as the sequence is bounded to begin with). That might resonate with someone else. It reminds me of a classical exercise where if you have a bounded function with bounded 2nd derivative, then the 1st derivative will approach $0$. $\endgroup$ Commented Sep 25, 2019 at 18:47
  • $\begingroup$ @RobertWolfe Good perspective! I am not sure that there is an analogous result in differential calculus. It seems that $f(x) = \sin(x)$ gives a counter-example for what you claimed in the last sentence. $\endgroup$ Commented Sep 26, 2019 at 0:18
  • $\begingroup$ @AoS you're right. I remembered it incorrectly. If the 2nd derivative approaches $0$, then the first does as well. If you have Baby Rudin laying around it's Ex. 15 in Chp. 5. $\endgroup$ Commented Sep 26, 2019 at 2:09

2 Answers 2

Reset to default
2
$\begingroup$

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 $|y_1-y_r|\leq 2c$. $\Box$

$\endgroup$
2
  • $\begingroup$ PS: I doubt I would have been able to construct this proof on an exam! $\endgroup$ Commented Sep 25, 2019 at 0:31
  • $\begingroup$ [Several months later]: And the reason for the recent unexplained downvote is... ? $\endgroup$ Commented Dec 10, 2019 at 14:27
0
$\begingroup$

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 $\forall m > N, \left| x_{n+1} - x_n \right| < \delta + \sqrt{2 \delta D} \leq 2\sqrt{2 \delta D} \leq \sqrt{8 \delta D} \leq \epsilon$

$\endgroup$
3
  • $\begingroup$ You lost me at the sentence "Furthermore, we know that $|(x_{n+2}−𝑥_{n+1})−(𝑥_{n+1}−x_n)|<\delta$, so we have that $|x_{𝑛+1}−𝑥_n|>(𝑘−1)\delta$." I wonder if there is some indexing error. [I think you are trying to use both $\epsilon$ and $\delta$, where in my proof I am using a limit to conclude an exact linear recurrence relation for limiting values $y_i-2y_{i+1} + y_{i+2}=0$, where in your proof it suffices to say that recurrence relation is at most $\delta$ away from zero. Then I think you want a relation between $\epsilon$ and $\delta$ similar to my size between $r$ and $\epsilon$.] $\endgroup$ Commented Sep 25, 2019 at 17:14
  • $\begingroup$ @Michael, thanks for catching my typo. I meant that since $\left| x_{n+1} - x_{n} \right| \geq k\delta$ and $\left| \left(x_{n+2} - x_{n+1}\right) - \left(x_{n+1} - x_{n}\right)\right| < \delta$, we have that $\left| x_{n+2} - x_{n+1} \right| > (k-1)\delta$. $\endgroup$ Commented Sep 25, 2019 at 18:32
  • $\begingroup$ I added an extra line to explain my reasoning. $\endgroup$ Commented Sep 25, 2019 at 18:42

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.