calculate radius of convergence Let $\{a_n\}_{n=0}^{\infty}$ be sequence such that
$$a_1 = a_0 = 1$$
$$a_{n+1}=a_n+ a_{n-1}$$
show that the radius of convergence of $\sum\limits_{n=0}^{\infty \:}a_nx^n$ is $\frac{-1+\sqrt{5}}{2}$.
Thanks!
 A: From the ratio test, the radius of convergence is
$$|x| < \big[\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}\big]^{-1} =L^{-1}$$
If the limit exists, then from the recursion
$$L - 1= \lim\frac{a_{n+1}}{a_n}-1=\lim\frac{a_{n-1}}{a_n}= L^{-1}$$
Then $L^{-1} = \frac{-1+ \sqrt{5}}{2}.$
To prove the limit exists, first establish the following identity for $m>n$:
$$a_ma_{n+1}-a_{m+1}a_{n}= (-1)^na_{m-n}.$$
Rearranging we get
$$\Big|\frac{a_{m+1}}{a_{m}}-\frac{a_{n+1}}{a_{n}}\Big| = \frac{a_{m-n}}{a_ma_n},$$
This can be used to establish that $a_{n+1}/a_n$ is a Cauchy sequence since the RHS converges to $0$ as $m,n \rightarrow \infty$.
A: You can find explicitly what a_n are: consider the next matrix A=[1,1;1,0] and notice (a_n+1, a_n)=A*(a_n,a_n-1)
From here you can by induction show: (a_n+1, a_n)=A^n *(a_1, a_0)
Now in order to calculate A^n use A=(I+B) where B=[0,1;0,0] and newton's binomial.
From here you can use caushy-hadamrd easily and finish.
A: The terms are Fibonacci numbers, and the ratio of consecutive ones is $\phi=\frac{1+\sqrt{5}}{2}$.  Now you can use the ratio test.
