Convergent sequence or not?

zebraman

1. The problem statement, all variables and given/known data
Let (x_n) be a real sequence which satisfies |x_n - x_(n+1)| < (1/n) for all natural numbers n.

Does (x_n) necessarily converge? Prove or provide counterexample.


2. Relevant equations
Cauchy Criterion for sequences


3. The attempt at a solution
I figured at first that this would be easily solved by determining if this sequence was a Cauchy sequence since the difference between the terms decreases with each successive term, but you don't know that you can always find a point after which the terms x_n, x_m have a difference of less than an epsilon. Any suggestions?

Sethric

Can you think of a series whose partials sums diverge while satisfying the requirement:
|s_n - s_(n+1)| < (1/n)?

zebraman

Well I was thinking the sequence of partial sums of the sequence x_n = 1/n diverges (Harmonic Series). But I guess since |x_n + x_(n+1)|<(1/n) that won't work.

Sethric

In the sequence

[tex] s_n = \sum_{i=1}^n \frac{1}{i}[/tex]

what is the value of |[tex]s_n - s_{n+1}[/tex]|?

zebraman

|s_n - s_(n+1)| = 1/(n+1), so are you saying that will serve as a counter argument? Because doesn't the sequence x_n = 1/n converge?

Dick

1/n converges. But sethric is suggesting using the sequence s_n, not 1/n. If you know the series 1/n diverges then you know the sequence of partial sums diverges. If you want a more easily expressed answer you might want to think about using an approximation to s_n. What is it?

zebraman

Sorry, I don't understand what you're asking.

Dick

I'm asking if you know that s_n is approximately equal to log(n) by an integral test.

zebraman

No, but is that important?

Sethric

Dick is correct, I was suggesting to use the sequence s_n. You have already shown:

|s_n - s_(n+1)| = 1/(n+1) < 1/n

You have also already said s_n diverges.

Dick

No, it's not important if you use sethric's suggestion. It's another series that has similar properties to the s_n.