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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.........

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  • $\begingroup$ Can you include how that answer $2 \over {1 + \sqrt{5}}$ came ? $\endgroup$ – Srinivas K Mar 6 '15 at 6:18
  • $\begingroup$ The solution to the recurrence will grow as $\phi^n$ (almost-)regardless of the initial conditions. $\endgroup$ – Gerry Myerson Mar 6 '15 at 6:33
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Mar 6 '15 at 6:36
  • $\begingroup$ 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.. $\endgroup$ – Empty Mar 6 '15 at 7:40
  • $\begingroup$ Duplicate: math.stackexchange.com/questions/106362/… $\endgroup$ – Marco Cantarini Mar 6 '15 at 8:31
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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.

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  • $\begingroup$ Please multiply and divide your answer by $\sqrt{5} -1$ and then, if you want, we can discuss. :) $\endgroup$ – incognito Mar 6 '15 at 14:15
  • $\begingroup$ OK..OK..I find it...It is clear to me now...I did this answer..but did not rationalizing .. $\endgroup$ – Empty Mar 6 '15 at 15:41

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