# Thread: Help with sequence counterexample

1. ## Help with sequence counterexample

Suppose that a sequence (xn) satisfies the following inequality:
abs(xn+1 - xn) < abs(xn - xn-1) for n = 2,3,4,.....

Does this sequence converge or is there a counterexample for it being divergent? If it converges, how would you prove it?

I would love some help on this particular problem. I'm not really sure if it converges or not. Thanks.

2. Originally Posted by PvtBillPilgrim
Suppose that a sequence (xn) satisfies the following inequality:
abs(xn+1 - xn) < abs(xn - xn-1) for n = 2,3,4,.....

Does this sequence converge or is there a counterexample for it being divergent? If it converges, how would you prove it?

I would love some help on this particular problem. I'm not really sure if it converges or not. Thanks.
Let $x_n = \sum_{i=1}^n \frac{1}{i}$.

3. Hello, PvtBillPilgrim!

Suppose that a sequence $x_n$ satisfies the following inequality:
. . $|x_{n+1} - x_n| \;< \;|x_n - x_{n-1}|$ . for $n \,=\,2,3,4,\cdots$

Does this sequence converge or is there a counterexample for it being divergent?
If it converges, how would you prove it?

TPH is absolutely correct.
. . Let me translate it . . .

It says: the difference of consecutive terms is decreasing.
. . The terms are "getting closer."
It would seem that the series converges.

But the Harmonic Series: . $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ . satisfies the inequality,
. . and it diverges.

We have found a counterexample.

Therefore, the sequence may or may not converge.

4. However if $|x_{n+1} - x_n | \leq k |x_n - x_{n+1}|$ where $0 then it will converge.