Consider the power series $$\sum_{n=0}^{\infty}a_nz^n.$$
where, $a_0=0$ , $a_1=1$ , $a_n=a_{n-1}+a_{n-2}$.
Find the radius of convergence of the power series.
MY Attempt :
Clearly $\{a_n\}$ is a Fibonacci sequence.
Let, $R$ be the radius of convergence of the power series.
We have , $$\frac{1}{R}=\lim_n\sup\left|\frac{a_{n+1}}{a_n}\right|$$
$$=\lim_n\sup\left|\frac{a_n+a_{n-1}}{a_n}\right|$$
$$1+\lim_n\sup\left|\frac{a_{n-1}}{a_n}\right|.$$
But I can't write $\lim_n\sup\left|\frac{a_{n-1}}{a_n}\right|$ in terms of $R$ such that we can find out $R$ by solving the equation involving $R$.
Again we know that the $n$-th term of Fibonacci sequence is $$a_n=\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right].$$
From this I find that the radius of convergence of the power series is $\frac{2}{1+\sqrt 5}$.
Is this answer correct ?
If NOT what is the correct answer ?
But I want to find the radius of convergence NOT using the $n$-th term of Fibonacci sequence. How I can find it ?
Please help…
Thanks in Advance………
Best Answer
Let us re-write the recurrence relation $$a_{n+1} = a_{n} +a_{n-1} $$ $$\frac{a_{n+1}} {a_{n}} = 1+\frac{a_{n-1}} {a_{n}} $$ Recursion allows us to write $$\frac{a_{n+1}} {a_{n}} = 1+\frac{1} {1+\frac{a_{n-2}} {a_{n-1}} } $$ One can keep repeating the same to get what is called as a continued fraction which in this case looks like: $$\frac{a_{n+1}} {a_{n}} = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}} $$ One should note that it will terminate at some point as we are supposed to get $\frac{a_0 }{a_1}$ at the n-th use of recurrence relation. But since we wish to find limit of the ratio $\frac{a_{n+1}} {a_{n}} $ as $n \rightarrow \infty$ we can write it as an infinite continued fraction. Now to find the value consider :$$\frac{1} {R} = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}} = 1 + \frac{1} {\frac{1} {R} } $$ Which yields the quadratic equation: $$R^2 + R - 1=0$$ The minimum root (absolute value) of which gives us the radios of convergence. I'll leave it to you.