# Complex power series : Radius of Convergence

Could anyone please suggest me how to deal with these questions (Complex Variable) : Note that all problems are in $\mathbb{C}$.

3.1 Determine the radius of convergence $\rho$ of each of the following series : $$\sum_{n=0}^\infty \frac{z^n}{n!}, \sum_{n=0}^\infty \frac{z^n}{n^2}, \sum_{n=0}^\infty \frac{z^n}{n}, \sum_{n=0}^\infty n!z^n$$ I use the ratio test and obtain the result : $\infty, 1, 1, 0$. , respectively.

The problem is : 3.2 For those with $\rho = \infty$, can you conclude anything about "convergence at infinity" ? I do not know what is "convergence at infinity". I search both in the text books and the internet already, but it seems no clues.

3.3 Find the radius of convergence of the power series : $$\sum a_nz^n,$$ where $a_0 = 0, a_1 = 1, a_n= a_{n-1} + a_{n-2}$ for all $n > 1$.

I do not know how to handle this series since $a_n$ is defined recursively, so most of method for determining radius of convergence cannot apply.

• When the radius of convergence is infinite the series defines an entire function. There are three cases of what can happen at $\infty$: (1) The function has a finite limit. Then by Liouville theorem it is a constant function. (2) The limit is $\infty$ itself. Then the function is a polynomial. In a sense, you can say that it converges to $\infty$ at $\infty$. (3) It has no limit, finite or infinite. This cases occurs if the series has infinite number of non-zero terms. – Pp.. Feb 5 '15 at 4:26
• For (3.3) Use that the radius of convergence goes all the way up to the first singularity of the function. Using the recurrence we can compute the function. Call $f(z):=\sum a_nz^n$. Multiply the recurrence by $z^n$ and add for all values of $n$. You get an equation for $f$. – Pp.. Feb 5 '15 at 4:33
• Thank you very much for your suggestion. I guess that the first comment is suggesting about "convergence at infinity", but I do not know about defining an entire function yet. Now, I know all basic algebraic property of $\mathbb{C}$. Also, the analytic equation-The Cauchy Reimann equation. Now I am studying complex power serie and radius of convergence. I am not sure I can use what is called Liouville thoerem. – user117375 Feb 5 '15 at 4:40
• For the second comment, I do not know now about singularities. Do you have a simpler notation in doing the problem ? I am sorry for my limited knowledge in complex variables. – user117375 Feb 5 '15 at 4:42

I think the solution is provided by Pp.. in his comments but if you didn't get it then first follow this link from Wikipedia which is about this series and its generating function. The second hint as provided by Pp.. is that you need to find the singularities of the generating function of this power series which is $f(z) = \frac{1}{1-z-z^2}.$ As is obvious the singularities are $\frac{\sqrt{5} \pm 1}{2}$. The minimum of which is $\frac{\sqrt{5} - 1}{2}$ and that is the radius of convergence.
I simplify it into this form : Let $P_N(z) = \sum_{n=1}^N a_nz^n$ for all $N \in \mathbb{N}$. Then let $$Q_N(z) = (z^2+z-1)P_N(z) = (z^2+z-1) \sum_{n=1}^N a_nz^n \\ = \sum_{n=0}^N a_nz^{n+2} + \sum_{n=0}^N a_nz^{n+1} - \sum_{n=0}^N a_nz^n \\ = -a_0 + (a_0 - a_1)z + (a_0 + a_1 - a_2)z^2 + (a_1 + a_2 - a_3)z^3 + ... + (a_{N-2} + a_{N-1} - a_N)z^N + a_Nz^{N+1} + a_{N-1}z^{N+1} + a_Nz^{N+2} \\ = -z + a_{N+1}z^{N+1} + a_Nz^{N+2}.$$