# Radius of convergence of a power serise involving the Fibonacci sequence.

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 $$\frac{2}{1+\sqrt 5}$$.

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 ?

• Can you include how that answer $2 \over {1 + \sqrt{5}}$ came ? Mar 6, 2015 at 6:18
• The solution to the recurrence will grow as $\phi^n$ (almost-)regardless of the initial conditions. Mar 6, 2015 at 6:33
• The answer you gave is correct. One could prove that the ratio $\frac{a_{n+1}}{a_n}$ has a limit, and then the fact it is $\frac{\sqrt{5}+1}{2}$ follows easily from the recurrence $a_n=a_{n-1}+a_{n-2}$. Just divide through by $a_{n+1}$. The fact that the limit exists is a bit unpleasant, but doable. It is definitely easier to work with the Binet formula that you quoted. Mar 6, 2015 at 6:36
• I tried in this way..but I can't understand that what is the limit of $\frac{a_{n-1}}{a_{n+1}}$ & the limit of $\frac{a_{n-2}}{a_{n+1}}$. If you more details to find the limits then it will helpful to me.. Mar 6, 2015 at 7:40
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.
• Please multiply and divide your answer by $\sqrt{5} -1$ and then, if you want, we can discuss. :) Mar 6, 2015 at 14:15