How to prove limit of a_{n+1}/a_n for a_{n+2}=2a_{n+1}+a_n w/ a_1/a_0 not eq to

[sktsasus]

I am a little confused as to how I can compute the limit for this.

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Problem 8. Let \(\displaystyle \, \left(a_n\right)_{n=0}^{\infty}\,\) be a sequence defined via the recursion

. . . . .\(\displaystyle a_{n+2}\, =\, 2\,a_{n+1}\, +\, a_n,\, \mbox{ for }\, n\, \geq\, 0\)

Assume that a0 and a1 are not both zero, and assume that

. . . . .\(\displaystyle \dfrac{a_1}{a_0}\, \neq\, 1\, -\, \sqrt{\strut 2\,}\)

Prove the following:

. . . . .\(\displaystyle \displaystyle \lim_{n\rightarrow \infty}\, \dfrac{a_{n+1}}{a_n}\, =\, 1\, +\, \sqrt{\strut 2\,}\)

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I know the sequence comes out to be x^2 = 2x + 1 but I am not sure how to proceed further.

Any help?

 

[ksdhart2]

I'd begin by "translating" the given recurrence relation for different indices, like so:

\(\displaystyle a_{n+2}=2a_{n+1}+a_{n}\)

\(\displaystyle a_{n+1}=2a_{n}+a_{n-1}\)

\(\displaystyle a_{n}=2a_{n-1}+a_{n-2}\)

Now plug those values into the limit:

\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \dfrac{a_{n+1}}{a_{n}} \right) = \lim_{n \to \infty} \left( \dfrac{2a_{n}+a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

Break this apart into two fractions:

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( \dfrac{2a_{n}}{2a_{n-1}+a_{n-2}} + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

But, hold on... we know what a_n equals, by using the recurrence relation. What happens if we plug that in?

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( \dfrac{2(2a_{n-1}+a_{n-2})}{2a_{n-1}+a_{n-2}} + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( 2 + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( 2 + \dfrac{a_{n-1}}{a_n} \right)\)

Hence, we have this new relation:

\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \dfrac{a_{n+1}}{a_{n}} \right) = 2 + \lim_{n \to \infty} \left( \dfrac{a_{n-1}}{a_n} \right)\)

Can you finish up from here?

 

[sktsasus]

I'd begin by "translating" the given recurrence relation for different indices, like so:

\(\displaystyle a_{n+2}=2a_{n+1}+a_{n}\)

\(\displaystyle a_{n+1}=2a_{n}+a_{n-1}\)

\(\displaystyle a_{n}=2a_{n-1}+a_{n-2}\)

Now plug those values into the limit:

\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \dfrac{a_{n+1}}{a_{n}} \right) = \lim_{n \to \infty} \left( \dfrac{2a_{n}+a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

Break this apart into two fractions:

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( \dfrac{2a_{n}}{2a_{n-1}+a_{n-2}} + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

But, hold on... we know what a_n equals, by using the recurrence relation. What happens if we plug that in?

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( \dfrac{2(2a_{n-1}+a_{n-2})}{2a_{n-1}+a_{n-2}} + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( 2 + \dfrac{a_{n-1}}{2a_{n-1}+a_{n-2}} \right)\)

\(\displaystyle \displaystyle = \lim_{n \to \infty} \left( 2 + \dfrac{a_{n-1}}{a_n} \right)\)

Hence, we have this new relation:

\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \dfrac{a_{n+1}}{a_{n}} \right) = 2 + \lim_{n \to \infty} \left( \dfrac{a_{n-1}}{a_n} \right)\)

Can you finish up from here?

@ksdhart2 Wow! Thank you so much for the detailed response! I completely follow what you have done. However, I have a question about the limit of the an-1/an. Since there aren't any number values attached to either of those two, how am I supposed to compute a limit. Are we supposed to derive the values from the given values of a1 and a0. But that too is a little unclear. If you could just clear this up, I think I will be able to completely solve it.

Thank you for the tremendous help so far!

 

[stapel]

I'd begin by "translating" the given recurrence relation for different indices....

Hence, we have this new relation:

\(\displaystyle \displaystyle \lim_{n \to \infty} \left( \dfrac{a_{n+1}}{a_{n}} \right) = 2 + \lim_{n \to \infty} \left( \dfrac{a_{n-1}}{a_n} \right)\)

@ksdhart2 Wow! Thank you so much for the detailed response! I completely follow what you have done. However, I have a question about the limit of the an-1/an. Since there aren't any number values attached to either of those two, how am I supposed to compute a limit?

Have you tried anything, starting from the last line (quoted above), which left you with maybe two more lines to do?

Hint: If we name as "x" the (unknown) value of the limit on the left-hand side of the equation you were given, what expression then stands for the (unknown) value of the limit on the right-hand side of the equation?

Hint: Quadratic Formula.

If you get stuck, please reply showing your thoughts and efforts. Thank you!

 

[sktsasus]

Have you tried anything, starting from the last line (quoted above), which left you with maybe two more lines to do?

Hint: If we name as "x" the (unknown) value of the limit on the left-hand side of the equation you were given, what expression then stands for the (unknown) value of the limit on the right-hand side of the equation?

Hint: Quadratic Formula.

If you get stuck, please reply showing your thoughts and efforts. Thank you!

Yes, I have managed to solve it now. I was still a little confused at that point so I thought it would be better to ask but I manged to figure it out. Yes, the quadratic formula was useful. Thank you!