3
$\begingroup$

I am trying to find a sequence such that $\lim_{n\to\infty} |a_n - a_{n+1}|=0$, but the $(a_n)$ diverges

I tried thinking something periodic might work like $\sin(2\pi n)$, but that is convergent sequence

Edit : Never mind Log[n] works great. Figured it out.

4 Answers 4

12

Try the sequence of the partial sums of the harmonic series, i.e.

$$a_n:=1+\frac{1}{2}+...+\frac{1}{n}$$

Here, $\,|a_{n+1}-a_n|=\frac{1}{n+1}\xrightarrow [n\to\infty]{}0\,$ but, as we know, the series diverges and thus, its partial

sums sequence diverges as well.

  • 0
    But that's the series, not the sequence.2012-10-07
  • 1
    @jak, take **the sequence** of the partial sums of *the series*! This is a sequence...2012-10-07
  • 0
    No it's okay. I figured it out2012-10-07
  • 3
    "No"?! What "no"?2012-10-07
  • 0
    I think OP means "Now"?2012-10-07
  • 0
    No I got a better answer2012-10-07
  • 1
    @jak: I wouldn't really call it "better". In fact, in some sense taking $\log n$ and taking $1+1/2+\cdots+1/n$ are the same.2012-10-07
  • 0
    What about sharing it, @jak? That way we *all* learn.2012-10-07
  • 0
    It's in my edit a few second after your answer. Don't worry, you'll still get accepted even though I don't feel the harmonic sequence diverges2012-10-07
  • 0
    Hehe...I didn't worry because of that, @jak: I don't need to. I just wanted to know what sequence you came up with which is "better", and in what sense, than mine. That's all. And what diverges is the sequence of partial sums of the harmonic series, and whether you feel it or not has no relevance here.2012-10-07
  • 1
    @jak Take a look at [this](http://en.wikipedia.org/wiki/Harmonic_series_(mathematics))2012-10-07
  • 1
    If you like more $\ln(n)$ because is not a series, it actually is :$\ln(n)= \sum_{k=1}^n (\ln(k)-\ln(k-1))$... But I agree that it is easier to prove that $\ln(n)$ is divergent than proving that the harmonic series is divergent....2012-10-07
  • 0
    @PatrickLi, that's the sum. I don't see how $a_n = {1,1/2,1/3,1/4,1/5, ..., 1/n} diverges2012-10-07
  • 1
    @jak: $a_n = 1+1/2+\cdots+1/n$, not $a_n=1/n$.2012-10-07
  • 0
    But I am asking the sequence here2012-10-07
  • 2
    I see you still don't get it, @jak: The series is the infinite one $$\sum_{n=1}^\infty\frac{1}{n}$$ This series is well-known to be divergent, which means that its **sequence** of partial sums $$\{1\,,\,1+\frac{1}{2}\,,\,1+\frac{1}{2}+\frac{1}{3}\,,...,1+\frac{1}{2}+...+ \frac{1}{n}\,,...\}$$ is divergent. Is this last **sequence** we all were talking about from the beginning!2012-10-07
  • 0
    @jak: That **is** a sequence. The fact that there are summations in there doesn't change anything.2012-10-07
  • 0
    @jak The sequence is $a_1=1$, $a_2=1+\frac{1}{2}$, $a_3=1+\frac{1}{2}+\frac{1}{3}$, ...2012-10-07
  • 0
    @PatrickLi and everyone. Is tihs what you guys meant? $$a_n = \left \{ \sum_{k=1}^{n} \frac{1}{k} \right \}_{n=1}^{\infty} $$2012-10-07
  • 1
    @jak: Yes. (15char15char15char)2012-10-07
  • 0
    Thanks, really nice examples.2012-10-07
  • 0
    Why didn't you guys write that in the first place then?! Sheeesh! Would've made more sense to me2012-10-07
  • 2
    Sheeesh!?? Don't you think it's weird we *all* knew what was going on and only you didn't? Perhaps if you said you don't know anything about series we *all* would have tried to make things easier to you...2012-10-07
7

or $a_n=\sqrt{n}$ also works well.

  • 0
    Do you think you could elaborate on that one?2012-10-07
  • 0
    @jak: What do you need clarification on? What is $\lim_{n\rightarrow\infty} (\sqrt{n}-\sqrt{n-1})$?2012-10-07
  • 0
    @wj32, how is that limit 0? Do you multiply that thing with the conjugate and that goes to 0? I haven't done the algebra, but that would be my guess2012-10-07
  • 4
    @jak : $\sqrt{n}-\sqrt{n-1} = \frac{(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})}{\sqrt{n}+\sqrt{n-1}}=\frac{1}{\sqrt{n}+\sqrt{n-1}} \to 0$ as $n\to \infty$2012-10-07
5

Or consider the sequence $\bigl\{0,{1\over2},1,{2\over3},{1\over3},0,{1\over4},{2\over4},{3\over4},1,{4\over5},\ldots\bigr\}$.

4

The sequence $a_n=\ln(n)$ has the desired property,

$$\lim_{n\to \infty } |\ln(n)-\ln(n+1)| = \lim_{n\to \infty } \left|\ln\left(\frac{n}{n+1}\right)\right| = 0 \,.$$

but, $\lim_{n\to \infty } \ln(n) =\infty\,. $