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.
$\endgroup$2 Answers
$\begingroup$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 $L$?
$\endgroup$ 7 $\begingroup$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 $b_n$.
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