Radius of convergence for $a_n=(a_{n-1}+a_{n-2})/2$ for $n \ge 2$.

Define a sequence $a_0,a_1,a_2,...$ by setting $a_0=1$, $a_1=2$, $a_n=(a_{n-1}+a_{n-2})/2$ for $n \ge 2$.

A) Find the radius of convergence of the series $\sum_{n=0}^{\infty}a_nz^n$.

B) Find an explicit formula for the function f(z) defined by the series of part A.

My approach for A

Using the definition of ratio test, I obtain $$\frac{|a_{n-1}+a_{n-2}|}{|1+a_{n-1}|}$$ So I didn't know what to do next as $n \to \infty$ and went ahead to plug in some numbers $n=2,3..$. I get $1.5$ then it decreases to $1.2222...$ and it keeps decreasing so I'm guessing that the radius of convergence will be $R=1.5$.

My approach for B

Computing the first five terms of $a_n$,

I get $$a_0=1$$

$$a_1=2$$

$$a_2=1.5$$

$$a_3=1.75$$

$$a_4=1.625$$ So that $$f(z)=1+2z+1.75z^2+1.75z^2+1.625z^4+...$$

Am I doing this correctly?

• Nope, your calculation of $a_n$ is wrong start at $a_3$. You should try showing $a_{2n}$ is increasing, $a_{2n+1}$ is decreasing and then $\lim_{n\to\infty} a_n$ exists. Once you get that, you have the answer for $(A)$. – achille hui May 2 '14 at 0:19
• @achillehui Why are u putting $a_{2n}$ instead of $a_n$? – User69127 May 2 '14 at 0:41
• The even and odd sub sequences are monotonic in opposite directions. One is increasing while the other is decreasing. – achille hui May 2 '14 at 0:47

There is a theory for recursive relations like this. The "characteristic polynomial" for the recursive relation given for $a_n$ is

$x^2 - x/2 - 1/2$

This factors into $x + 1/2$ and $x - 1$, which breaks $a_n$ up into two parts: $a_n = b_n + c_n$, where $b_n$ satisfies $b_{n+1} = - b_n / 2$ and $c_{n+1} = c_n$. So if $b$ and $c$ are the first terms of these two sequences, then the sum of $a_n * z^n$ is the sum of two series,

$b ( 1 - z/2 + z^2 / 4 - ...) = b / ( 1 + z/2)$ which converges for $|z| < 2$, and

$c ( 1 + z + z^2 + ...) = c / (1 - z)$ which converges for $|z| < 1$.

So the original series converges for $|z| < 1$ to the function $b/(1+z/2) + c/(1-z)$, where

$b + c = a_0$and $-b/2 + c = a_1$

• Ah I sort of see what you're doing here. Are you saying I should use $2a_n-a_{n-1}-a_{n-2}=0$? – User69127 May 2 '14 at 0:46
• Whether you divide by 2 doesn't really matter. The main point is that factors of the characteristic polynomial correspond to the sequence being broken up into parts each of which satisfies a simpler recursive equation. – berkeleychocolate May 2 '14 at 4:31