A sequence $c_n$ is defined by the following recursion $c_{n+1} = c_n + c_{n-1}$ for every $n \geq 1$ and $c_0 = 1, c_1 = 2$.

-Let $a_n = \frac{c_{n+1}}{c_n}$, for every $n\geq 0$ and prove that $a_n = 1 + \frac{1}{a_{n-1}}$ for every $n \geq 1$.

-Calculate the limit of the following sequence $\frac{c_1}{c_0}, \frac{c_2}{c_1}, \frac{c_3}{c_2},...,\frac{c_{n+1}}{c_n}$ using the fact that this sequence is convergent

I tried close to everything to obtain the formula mentioned above but I did not succeed. What I obtained is $a_n=1 + \frac{c_{n-1}}{c_n}$ which is obviously wrong. As for calculating the limit, I don't have a clue what is meant by 'using the fact that this sequence is convergent'. Could anyone please help me out? Thank you in advance.


$C_{n+1}= C_n + C_{n-1}, A_n=\frac{C_{n+1}}{C_n}$

Substituting the first equation into the second equation you get: $A_n = 1 + \frac{C_{n-1}}{C_n}$.

From the second equation we can write: $C_{n-1} = \frac{C_n}{A_{n-1}}$.

Then we substitute this in our third equation and we get: $A_n= 1 + (\frac{C_n}{A_{n-1}})/C_n $.

is equal to $1 + \frac{1}{A_{n-1}}$.

  • $\begingroup$ pretty much like mine isn't it? and please use $\LaTeX$ $\endgroup$ – Dominic Michaelis Feb 16 '13 at 10:06
  • $\begingroup$ I was writing it before you posted your answer.. $\endgroup$ – Nedellyzer Feb 16 '13 at 10:13
  • $\begingroup$ I thought something like that, seems like he enjoys your answer more ... $\endgroup$ – Dominic Michaelis Feb 16 '13 at 10:15

If you want to calculate the limit use $$\lim_{n\rightarrow \infty} a_{n+1} = \lim_{n\rightarrow \infty} a_n$$ This formula is only true if the limit exists. Oh sry the Calculation: $$a_n= \frac{c_{n+1}}{c_n} = \frac{c_n +c_{n-1}}{c_n}= \frac{c_n}{c_n} + \frac{c_{n-1}}{c_n}=1+\frac{1}{a_{n-1}}$$ Maybe a small explanation for my hint, if the limits are equal we can say the limit is $a$, so $$a=1+\frac{1}{a}$$. Just solve this one (you get 2 solutions, you have to check which one is really the limit here your starting points help).


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