# Finding limit of a sequence defined by another sequence

Let there be $\{a_n\}$ such that $a_{n+2}=2a_{n+1}+2a_n$ and $\{b_n\}$ such that $b_n=\frac{a_{n+1}}{a_n}$ if $\lim_{n \to \infty}b_n=L$ what is the limit $L$?

I thought about if $L$ exist so I can use the arithmetic laws of limits.

$\lim b_n=\lim \frac{a_{n+1}}{a_n}$ and $\lim b_n=\lim \frac{2a_{n+1}+a_n}{a_{n+1}}$.

But could not find how to proceed to find the answer.

From $a_{n+2} = 2 a_{n+1} + 2 a_n$ it follows that $b_{n+1} = 2 + 2/b_n$. If $\lim_{n \to \infty} b_n = L$ exists, what can you conclude about $L$?
• I get $L_1=1-\sqrt{3},$ ,$L_2=1+\sqrt{3}$ can they be both? – gbox Nov 6 '15 at 8:56
• Both are used in a formula for $a_n$. As in every forward power iteration, the quotient converges to the larger of the two (largest of all eigenvalues), $L_2$, with a geometric rate of $L_1/L_2$. – LutzL Nov 6 '15 at 8:59
• @gbox: Can the limit of $(b_n)$ be negative? – Martin R Nov 6 '15 at 9:00
• You get this special result by following the answer of Michael, $\frac{a_{n+1}}{a_n}=\frac{Aλ^{n+1}+Bμ^{n+1}}{Aλ^n+Bμ^n}$. You get the general interpretation by writing the recursion in matrix form for the vectors $\binom{a_n}{a_{n-1}}$. – LutzL Nov 6 '15 at 9:09
First solve the recursion for $a_n$. The general solution is $a_n=A\lambda^n+B\mu^n$. Use the recursion to find a formula that $\lambda$ and $\mu$ satisfy; then calculate $b_n$.