Proving the Convergence of a Recursive Sequence

[Linday12]

Homework Statement

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

The Attempt at a Solution

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

Now, observe that a_{1}=1 < 3. If a_{k} < 3, then a_{k+1}=\sqrt{6+a_k} < \sqrt{6+3}, so a_{n} < 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.

 

[tiny-tim]

Hi Linday12!

(have a square-root: √ and try using the X₂ tag just above the Reply box )

Now, observe that a_{1}=1 < 3. If a_{k} < 3, then a_{k+1}=\sqrt{6+a_k} < \sqrt{6+3}, so a_{n} < 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 a_n < 3, then a_n+1 < 3 …

what is worrying you about that? ​

 

[Linday12]

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 a_k < 3, then a_k+1 = \sqrt{6+a_{k}} < \sqrt{6+3}, so a_n < 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.

 

[rs1n]

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 a_k < 3, then a_k+1 = \sqrt{6+a_{k}} < \sqrt{6+3}, so a_n < 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.

What you have just proved is: (\star) the validity of the statement "a_k&lt;3" implies the validity of the statement "a_{k+1}&lt;3." What this allows you to do is essentially re-use (\star) over and over. We know that a_1&lt;3. By (\star), the validity of a_1 &lt; 3 implies the validity of a_2 &lt; 3. But, again by (\star), the validity of a_2&lt;3 implies the validity of a_3&lt;3 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.

 

[tiny-tim]

Hi Linday12!

(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 a_n 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 has said, and try to convince yourself that it makes sense (it does! ).

Then try to apply it in other examples (such as a_n greater than 3 in this case).

 

[Linday12]

Thank you both very much! Those two posts cleared up my problems with it. I was thinking of it in an accumulating way (I have no idea why, series and sequence mix up I guess ), so now it makes sense.

Thanks again!