Quick Sequence Question

[ardentmed]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:

I'm mostly concerned about b and c at the moment. I got 0 for a after taking the limit, so it must be convergent. As for b, I got indeterminate form, so I used L'Hopital's Rule and got convergent (since it's a non-infinite value), but I'm doubting my answer.

As for c, I'm having a hard time proving the question. Do I just state that an > a(n+1) and substitute the given value for an and compare it with the next term in the sequence?

Thanks in advance.

 

[HallsofIvy]

Yes, \(\displaystyle \lim_{n\to\infty} \frac{n!}{(n+2)!}= \lim_{n\to\infty} \frac{1}{(n+1)(n+2)}\) which obviously converges to 0.

For (b) do you know the basic "property of limits", \(\displaystyle \lim_{n\to\infty} (a_n+ b_n)= \lim_{n\to\infty} a_n+ \lim_{n\to\infty} b_n\)? Of course, you want to take \(\displaystyle a_n= 3\) and \(\displaystyle b_n= \frac{(-1)^n}{n}\). Both of those limits should be easy.

For (c) there is a theorem that says that if a sequence is increasing and has an upper limit, the it converges.
To show that those are true, suppose, temporarily, that the sequence does converge to, say, A. Taking the limit on both sides of \(\displaystyle a_{n+1}= \sqrt{2+ a_n}\) (using the fact that \(\displaystyle \sqrt{2+ x}\) is continuous function) we get \(\displaystyle A= \sqrt{2+ A}\). Solving that for A will tell you what the limit should be. Use that to show that the sequence has an upper bound and that it is increasing. (I would try "proof by induction.)

 

[ardentmed]

Yes, \(\displaystyle \lim_{n\to\infty} \frac{n!}{(n+2)!}= \lim_{n\to\infty} \frac{1}{(n+1)(n+2)}\) which obviously converges to 0.

For (b) do you know the basic "property of limits", \(\displaystyle \lim_{n\to\infty} (a_n+ b_n)= \lim_{n\to\infty} a_n+ \lim_{n\to\infty} b_n\)? Of course, you want to take \(\displaystyle a_n= 3\) and \(\displaystyle b_n= \frac{(-1)^n}{n}\). Both of those limits should be easy.

For (c) there is a theorem that says that if a sequence is increasing and has an upper limit, the it converges.
To show that those are true, suppose, temporarily, that the sequence does converge to, say, A. Taking the limit on both sides of \(\displaystyle a_{n+1}= \sqrt{2+ a_n}\) (using the fact that \(\displaystyle \sqrt{2+ x}\) is continuous function) we get \(\displaystyle A= \sqrt{2+ A}\). Solving that for A will tell you what the limit should be. Use that to show that the sequence has an upper bound and that it is increasing. (I would try "proof by induction.)

Thanks a ton. I really appreciate the help.