Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 11 down vote accepted

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


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.

share|cite|improve this answer
@jak, take the sequence of the partial sums of the series! This is a sequence... – DonAntonio Oct 7 '12 at 4:05
"No"?! What "no"? – DonAntonio Oct 7 '12 at 4:07
@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. – wj32 Oct 7 '12 at 4:10
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! – DonAntonio Oct 7 '12 at 4:47
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... – DonAntonio Oct 7 '12 at 12:20

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

share|cite|improve this answer
Do you think you could elaborate on that one? – Hawk Oct 7 '12 at 5:04
@jak: What do you need clarification on? What is $\lim_{n\rightarrow\infty} (\sqrt{n}-\sqrt{n-1})$? – wj32 Oct 7 '12 at 5:08
@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 guess – Hawk Oct 7 '12 at 5:32
@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 $n\to \infty$ – Patrick Li Oct 7 '12 at 5:39

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

share|cite|improve this answer

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\,. $

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.