# Help with sequence counterexample

• Oct 25th 2007, 08:24 AM
PvtBillPilgrim
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.
• Oct 25th 2007, 08:34 AM
ThePerfectHacker
Quote:

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 $\displaystyle x_n = \sum_{i=1}^n \frac{1}{i}$.
• Oct 25th 2007, 08:48 AM
Soroban
Hello, PvtBillPilgrim!

Quote:

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

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

We have found a counterexample.

Therefore, the sequence may or may not converge.

• Oct 25th 2007, 01:42 PM
ThePerfectHacker
However if $\displaystyle |x_{n+1} - x_n | \leq k |x_n - x_{n+1}|$ where $\displaystyle 0<k<1$ then it will converge.