• Support PF! Buy your school textbooks, materials and every day products Here!

Sequences, induction help

  • Thread starter Linday12
  • Start date
  • #1
54
0

Homework Statement


Let a_n be defined recursively by
[tex]a_{1}[/tex]=1, [tex]a_{n+1}[/tex]=sqrt(6+[tex]a_{n}[/tex]) (n=1,2,3,...).
Show that lim n->infinity [tex]a_{n}[/tex] exists and find its value

The Attempt at a Solution


Observe that [tex]a_{2}[/tex]=[tex]\sqrt{6+1}[/tex]=[tex]\sqrt{7}[/tex] > [tex]a_{1}[/tex]. If [tex]a_{k+1}[/tex] > [tex]a_{k}[/tex], then [tex]a_{k+2}[/tex] = [tex]\sqrt{6+a_{k+1}}[/tex] > [tex]\sqrt{6+a_k}[/tex] = [tex]a_{k+1}[/tex], so {[tex]a_{n}[/tex]} is increasing by induction. I get that part, its the next part I'm a little confused about:

Now, observe that [tex]a_{1}[/tex]=1 < 3. If [tex]a_{k}[/tex] < 3, then [tex]a_{k+1}[/tex]=[tex]\sqrt{6+a_k}[/tex] < [tex]\sqrt{6+3}[/tex], so [tex]a_{n}[/tex] < 3 for every n by induction. How does this show that for all n, the value will be less than 3? Sorry, I'm not really sure how it shows that if you keep going on in the sequence it won't go past 3 eventually.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi Linday12! :wink:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
Now, observe that [tex]a_{1}[/tex]=1 < 3. If [tex]a_{k}[/tex] < 3, then [tex]a_{k+1}[/tex]=[tex]\sqrt{6+a_k}[/tex] < [tex]\sqrt{6+3}[/tex], so [tex]a_{n}[/tex] < 3 for every n by induction. How does this show that for all n, the value will be less than 3? Sorry, I'm not really sure how it shows that if you keep going on in the sequence it won't go past 3 eventually.
But you've proved it!

If an < 3, then an+1 < 3 …

what is worrying you about that? :smile:
 
  • #3
54
0
Sorry, I wasn't clear enough. I'm just not seeing how it shows that. I've memorized it and am not really worried about getting it wrong, but it's really not helpful if I don't understand that it keeps under the limit. I guess my induction skills are lacking.

Ok, so if ak < 3, then ak+1 = [tex]\sqrt{6+a_{k}}[/tex] < [tex]\sqrt{6+3}[/tex], so an < 3 for every n by induction.

What I'm not seeing is how this shows that it is below 3 for n. For example, as you keep going up to something like a100, how do you know that it won't add up to enough to push over 3. My (wrongful) reasoning is that if you keep adding more and more, it finally accumulates enough to surpass 3 and go to infinity. I can't seem to see how it approaches the limit. :frown:
 
  • #4
179
2
Sorry, I wasn't clear enough. I'm just not seeing how it shows that. I've memorized it and am not really worried about getting it wrong, but it's really not helpful if I don't understand that it keeps under the limit. I guess my induction skills are lacking.

Ok, so if ak < 3, then ak+1 = [tex]\sqrt{6+a_{k}}[/tex] < [tex]\sqrt{6+3}[/tex], so an < 3 for every n by induction.

What I'm not seeing is how this shows that it is below 3 for n. For example, as you keep going up to something like a100, how do you know that it won't add up to enough to push over 3. My (wrongful) reasoning is that if you keep adding more and more, it finally accumulates enough to surpass 3 and go to infinity. I can't seem to see how it approaches the limit. :frown:
What you have just proved is: [itex](\star)[/itex] the validity of the statement "[itex]a_k<3[/itex]" implies the validity of the statement "[itex]a_{k+1}<3[/itex]." What this allows you to do is essentially re-use [itex](\star)[/itex] over and over. We know that [itex]a_1<3[/itex]. By [itex](\star)[/itex], the validity of [itex]a_1 < 3[/itex] implies the validity of [itex]a_2 < 3[/itex]. But, again by [itex](\star)[/itex], the validity of [itex]a_2<3[/itex] implies the validity of [itex]a_3<3[/itex] and so on.

In induction proofs, what you're proving is that the validity of the k-th case implies the validity of the (k+1)-th case. Along with the base case, this allows you to conclude the statement is true for all values of k.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi Linday12! :smile:

(just got up :zzz: …)
My (wrongful) reasoning is that if you keep adding more and more, it finally accumulates enough to surpass 3 and go to infinity.
Nothing is accumulating, there's no sum, each an is on its own.
… I guess my induction skills are lacking. …
Yes, you need to convince yourself that induction is valid.

Go over what rs1n :smile: has said, and try to convince yourself that it makes sense (it does!! :biggrin:).

Then try to apply it in other examples (such as an greater than 3 in this case). :smile:
 
  • #6
54
0
Thank you both very much! Those two posts cleared up my problems with it. I was thinking of it in an accumulating way :uhh: (I have no idea why, series and sequence mix up I guess :blushing:), so now it makes sense.

:cool: Thanks again!
 

Related Threads on Sequences, induction help

Replies
13
Views
2K
  • Last Post
Replies
2
Views
864
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
821
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
2
Views
2K
Replies
11
Views
19K
Replies
1
Views
1K
Replies
3
Views
1K
Top