# Test Review 3 - convergence of series

1. Oct 15, 2005

### cmurphy

Problem: Suppose 0 < sn < 2 and sn+1 = root (sn + 2) for n in N. Prove
0 < sn < sn+1 < 2 holds for all n in N. Does sn converge? If so, what is the limit.

I am able to show that sn+1 < 2 by squaring the equation sn+1 = root (sn + 2) and making a substitution.

How would I go about showing that sn < sn+1?

Also, if sn < sn+1 < 2 for all n, then the series sn must converge (because it is bounded). In order to find the limit, could I take the limit of both sides of the equation (sn+1)^2 = sn + 2?

i.e. lim (sn+1)^2 = lim (sn+2)).
Let s = lim sn = lim sn+1
Then s^2 = s + 2
s^2 - s - 2 = 0
But then s = 2 and s = -1, which would indicate that there is no limit.

So which is it - is there a limit or not? (I thought that if a sequence is bounded, it must have a limit. Is that correct?)

2. Oct 15, 2005

### HallsofIvy

Staff Emeritus
Does the problem really say 0< sn< 2? If so that makes proving that 0< sn+1< 2 trivial! Are you sure it doesn't say 0< s1< 2?
What substitution, exactly? (And don't forget the 0< sn part!)
Hint: what are the roots of the equation x2- x- 2= 0?
and is an increasing sequence!
An increasing sequence with an upper bound must have a limit. That's part of the "monotone convergence" property.
Can you explain why "s= 2 and s= -1" would "indicate that there is no limit"?
(Notice that $x_n= \frac{n-1}{n}$ has the property xn< 1 for all n. What is its limit?)

3. Oct 15, 2005

### cmurphy

Yes, the problem does say that 0 < sn <2, and that sn+1 = root(sn +2).

Thus I said that (sn+1)^2 = sn + 2.
Then sn = (sn+1)^2 - 2

Thus 0 < sn = (sn+1)^2 - 2 < 2
And 2 < (sn+1)^2 < 4
So root 2 < sn+1 < 2.

Thus I have that sn+1 < 2.

I obviously have that 0 < sn, from the given statement. I cannot figure out how to show that sn < sn+1 to complete this part of the proof.

I did figure out that since the roots (limits) of the equation s^2 = s + 2 are 2 and -1, that the limit cannot be -1 since all sn > 0. Thus the limit is 2.

And, since the sequence is shown to be monotone and bounded, we know that it does need to converge.