Convergence radius of power series $\sum_{n=0}^\infty c_nz^n $ where $c_0 = 0$, $c_1 = 1$, $c_n =\frac{c_{n-1} + c_{n-2}}{2}$

Question

In my homework I'm supposed to find the convergence radius of the power set in the title. I'm allowed to assume $C = \lim_{n\to\infty} {c_{n+1}\over c_n}$ exists and we should try to figure out that $C$ has to be. I found $${c_{n+1}\over {c_n}} = {1\over 2} + {c_{n-1}\over{2c_n}} $$ so for $n\to\infty$: $$C = {1\over2} + {1\over{2C}} \Leftrightarrow C=1$$

which doesn't really help me further. Investigating $c_n - c_{n-1}$ didn't help me out either.

Any tips on what I might be missing here? I know how to find the radius once I've found the limit of the sequence, but can't seem to find it.

Every $c_n$ is the average of the two previous ones. So it's not larger than the larger of those, and not smaller than the smaller of them. Can you find a short interval that contains all $c_n$ for $n > 0$? — Daniel Fischer, Jan 22 '18 at 21:42
It is identical to https://math.stackexchange.com/questions/2559687/show-that-the-sequence-a-1-1-a-2-2-a-n2-a-n1a-n-2-converges-b/2559826#2559826 just different initial values ... try the methods there — rtybase, Jan 22 '18 at 21:43
Of course knowing that $c_{n+1}/c_n\to1$ completely determines the radius of convergence of the series... — Did, Jan 22 '18 at 21:44
Possible duplicate of Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy — Guy Fsone, Jan 22 '18 at 21:46
There are inequalities which relate $c_{n+1}/c_n$ with $\sqrt[n]{c_n}$ when you consider their liminfs and limsups. — , Jan 22 '18 at 21:52

Mostafa Ayaz · Accepted Answer · 2018-01-22T22:11:09.953

Assume at the most general case for a sequence $c_n$ we have a recursive equation as following:$$c_n=\sum_{i=1}^{k}a_ic_{n-i}$$. Now define $C(z)=\sum_{n=0}^{\infty}c_nz^{-n}$. This transformation is also referred to as z-transform. By taking z-transform from both sides of the recursive equation we arrive at the following equation:$$C(z)=C(z)\sum_{i=1}^{k}a_iz^{-i}$$which leads to the $characteristic$ $equation$ as below:$$\sum_{n=1}^{k}a_n\lambda^{-n}=0$$ and the $c_n$ itself can be described as a linear combination of eigenfunctions $\lambda_i^n$ as following:$$c_n=\sum_{i=1}^{k}b_i\lambda_{i}^n$$where all the $\lambda_i$'s are assumed to be simple roots of the characteristic function. Embarking on this, the characteristic equation of $2c_n=c_{n-1}+c_{n-2}$ is $2\lambda^2=\lambda+1$ which has two roots $\lambda=1$ and $\lambda=-\frac{1}{2}$ and leads us to $c_n=A(1)^n+B(-\frac{1}{2})^n$. Also $c_0=0$ and $c_1=1$ therefore $c_n=\frac{2}{3}-\frac{2}{3}(-\frac{1}{2})^n$ and by substitution we have:$$\sum_{n=0}^{\infty}c_nz^n=\sum_{n=0}^{\infty}\frac{2}{3}z^n-\frac{2}{3}(-\frac{1}{2}z)^n$$Convergence of the series requires $|z|<1$ and $|-\frac{1}{2}z|<1$ which finally leads to $|z|<1$ and $R=1$ therefore:$$\sum_{n=0}^{\infty}c_nz^n=\frac{z}{(1-z)(1+\frac{1}{2}z)}\quad,\quad |z|<1$$

Yep ... better, but still not enough, now we can see that $\lim\limits_{n\rightarrow\infty}c_n=\frac{2}{3}$ which means $\lim\limits_{n\rightarrow\infty}\left|\frac{c_n}{c_{n+1}}\right|=1$ which means that radius of converges is $1$ — rtybase, Jan 22 '18 at 22:15
And you answered the OP's question and a well deserved (+1) :) — rtybase, Jan 22 '18 at 22:30

Convergence radius of power series $\sum_{n=0}^\infty c_nz^n $ where $c_0 = 0$, $c_1 = 1$, $c_n =\frac{c_{n-1} + c_{n-2}}{2}$

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