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I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series). But then I thought "is it possible for the limit of $x_n$ to not exist at all?" and I think the answer is no,because I couldn't find any examples, but I am not sure. Could you please tell me what your opinion is on this?

Thanks in advance!

Britanica
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3 Answers3

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Take $$x_n=\sin(\sqrt{n})$$

By MVT,

$$\lim_{n\to +\infty}(\sin(\sqrt{n+1})-\sin(\sqrt{n}))=0$$

but

$$\lim_{n\to+\infty}x_n$$ does not exist.

OR

$$x_n=\sum_{k=0}^n\frac{1}{k!}$$ as a sequence of rationnals.

it has no limit in $ \Bbb Q$.

hamam_Abdallah
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    A slightly easier way to get $ \lim_{n\to +\infty}(\sin(\sqrt{n+1})-\sin(\sqrt{n}))=0$ for those who have not seen the MVT is to use $\sin(a)-\sin(b) = 2\sin(\frac{1}{2}(a-b))\cos(\frac{1}{2}(a+b)) \in [- 2\sin(\frac{1}{2}(a-b)), 2\sin(\frac{1}{2}(a-b))] $ –  Jan 06 '21 at 20:49
  • @oskarszarowicz Thank you very very much. – hamam_Abdallah Jan 06 '21 at 21:20
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Let $$x_n-x_{n-1}=\frac{(-1)^{\lceil\log_2n\rceil}}n$$ This is in effect taking the harmonic $x_n=H_n$ example, which has $\infty$ as a limit, and "folding it" into infinitely many intervals where $x_n$ goes up and down alternately by at least $\frac12$ per interval:

  • From $x_2$ to $x_4$, $\frac13+\frac14>\frac12$ is added
  • From $x_4$ to $x_8$, $\frac15+\frac16+\frac17+\frac18>\frac12$ is subtracted, etc.

Thus $x_n$ cannot have a limit.

Parcly Taxel
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The answer is yes!

Not only you can find examples were the limit is not finite but is for example $\infty$. You can also have examples were a full interval is a set of limit points of the sequence.

Look at a sequence that will slide back and forth from $0$ to $1$ with steps decreasing to zero for example. Let's for example consider $\{x_n\}$ whose initial terms are $$ \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \dots$$

Formal definition of $\{x_n\}$ is

$$ x_n=\begin{cases} \frac{1}{2} &\text{ for } n= 1\\ \frac{n}{k+2} - \frac{k(k+1)}{2(k+2)} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2} \end{cases}$$

The values of the sequence $\{x_n\}$ are in $(0,1) \cap \mathbb Q$. Moreover, one can notice that $\{x_n\}$ takes each rational number of $(0,1)$ as value an infinite number of times. Indeed for $\frac{p}{q} \in (0,1)$ with $1 \le p \lt q$ and $m \ge 1$ we have $$ \frac{(mq-2)(mq-1)}{2} \lt \frac{(mq-2)(mq-1)}{2} + mp$$ and $$\begin{aligned} \frac{(mq-2)(mq-1)}{2} + mp &\le \frac{(mq-2)(mq-1)}{2} + m(q-1)\\ &\le \frac{(mq-2)(mq-1)}{2} + mq-1\\ &= \frac{(mq-1)mq}{2} \end{aligned}$$ Hence $$\begin{aligned} r_{\frac{(mq-2)(mq-1)}{2} + mp} &= \frac{(mq-2)(mq-1)}{2mq} + \frac{mp}{mq} - \frac{(mq-2)(mq-1)}{2mq}\\ &= \frac{mp}{mq}\\ &= \frac{p}{q} \end{aligned}$$ proving the desired result. As the rational numbers of the segment $(0,1)$ are dense in $[0,1]$, we can conclude that the set of limit points of $\{x_n\}$ is exactly the interval $[0,1]$.

See here for additional examples.

mathcounterexamples.net
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