hello all
this is my basically my linear algebra approach enjoy, I don't know if anybody knew this but there is a general formulae in matrix form
A^n=\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)
where
A=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)
by solving the characteristics equation
|A-\lambda I|=0
we get
(1-\lambda)(-\lambda)-1=0
\lambda^2-\lambda-1=0
as a result the eigenvalues are
\lambda=\phi;(1-\phi)
then by multiplying by \lambda^n
we get
\lambda^{n+2}-\lambda^{n+1}-\lambda^{n}=0
and so
n\mapsto \phi^n
also
n\mapsto (1-\phi)^n
just to note the eigenvectors are linear independent of each other and so the eigenvectors can be put into a linear combination, as a result we can have
n\mapsto a\phi^n+b(1-\phi)^n
to accommodate for all linear combinations and by adjusting the coefficients for the initial values we get
F_n=\frac{\phi^n+(1-\phi)^n}{\sqrt{5}}
now we let
a=\phi
and
b=1-\phi
a-b=\sqrt{5}
then
F_n=\frac{a^n-b^n}{a-b}
then we have
\frac{F_{n+1}}{F_n}=\frac{a^{n+1}-b^{n+1}}{a^n-b^n}
we then take the limit of both sides and we get
\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_n}=\lim_{n\rightarrow\infty}\frac{a^{n+1}-b^{n+1}}{a^n-b^n}
since
|b|<1
and
\lim_{n\rightarrow\infty}b^n=0
then
\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_n}=\lim_{n\rightarrow\infty}\frac{a^{n+1}}{a^n}=a=\phi
oh yes about the existence of the limit that aint too hard all i did was show that
1\le\frac{F_{n+1}}{F_n}\le 2
by using the fact
a_{n+1}=1+\frac{1}{a_n}
then i let
a_n=\frac{F_{n+1}}{F_n}
and showed that the subsequence (by mathematical Induction)
a_{2n+1} is increasing and
a_{2n} is decreasing
and so since all terms are positive then both subsequences converge to the same limit
steven