Test Review 3 - convergence of series

[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?)

 

[HallsofIvy]

Problem: Suppose 0 < sn < 2 and sn+1 = root (sn + 2) for n in N.

Does the problem really say 0< s_n< 2? If so that makes proving that 0< s_n+1< 2 trivial! Are you sure it doesn't say 0< s₁< 2?

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.

What substitution, exactly? (And don't forget the 0< s_n part!)

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

Hint: what are the roots of the equation x²- x- 2= 0?

Also, if sn < sn+1 < 2 for all n, then the series sn must converge (because it is bounded).

and is an increasing sequence!

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?)

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 x_n< 1 for all n. What is its limit?)

 

[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.

So the only part left is showing that sn < sn+1. Please help!

 

[siddharth]

To show sn<sn+1 use the relation between sn and sn+1.
That is, showing sn<sn+1 is the same as showing sn<root(sn+2).