# Why does the radius of convergence stay the same when adding two power series with the same radius of convergence?

If we build the sum of two power series , $\sum a_kz^k$, $\sum b_kz^k$ both with radius of convergence 1, why does the resulting power series: $$\sum (a_k+b_k)z^k$$ also have RoC 1?

Is it because : $$|\frac{a_k}{a_{k+1}}| = 1 , |\frac{b_k}{b_{k+1}}|=1$$

$$|\frac{a_k+b_k}{a_{k+1}+b_{k+1}}|$$

that doesnt really do anything T_T .... If you see how to show this, please do tell.

-
Not true. The series has radius of convergence at least 1, but it may very well be larger. –  Andres Caicedo Sep 13 '12 at 0:18

Let $p(z)$ have radius of convergence $1$, and let $q(z)$ have radius of convergence $4$. Then $(q+p)(z)$ and $(q-p)(z)$ have radius of convergence $1$, but their sum has radius of convergence $4$.
For any $k\ge 1$, including "$k=\infty$", we can find power series with radius of convergence $1$ whose sum has radius of convergence $k$.
I dont see, why cant the sum be smaller than one? And how would you create one with k= infinity if both are required to have the same radius of convergence (by setting $b_k = -a_k$ ? ) –  bakabakabaka Sep 13 '12 at 0:32
@bakabakabaka: Sure, $b_k=-a_k$ is good. For for a fancier example, $e^z-\frac{1}{1-z}$ and $e^z+\frac{1}{1-z}$. As for your question about why not less than $1$, the sum has radius of convergence $\ge 1$ by say the ratio test. –  André Nicolas Sep 13 '12 at 0:39
@bakabakabaka: Formally, use Ratio Test or Root Test. More informally, if $p(z)$ is nice only when $|z|\lt 1$, and $q(z)$ is nice for $|z|\lt 4$, then $q(z)+p(z)$ is nice only when $|z|\lt 1$. To put it another way, $p(z)$ has a singularity somewhere on the circle $|z|=1$, and $q(z)$ is OK for any $z$ with $|z|\lt 4$, so $q(z)+p(z)$ goes bad somewhere on $|z|=1$. –  André Nicolas Sep 13 '12 at 1:17
@bakabakabaka: Informal argument for $q-p$ is the same. For formal, Ratio Test on coefficients is unwieldy, so here is another way. Let $\sum p_n z^n$ have radius of convergence $1$, let $\sum q_n z^n$ have radius of convergence $4$. So for example $\sum p_n z^n$ diverges if $|z|=1.01$. But $\sum q_n z^n$ converges for $|z|=1.01$. It follows that $\sum (p_n+q_n)z^n$ diverges for $|z|=1.01$. So radius of convergence of sum is $\le 1.01$. In same way, it is $\le 1+\epsilon$ for every positive $\epsilon$, so it is $\le 1$. –  André Nicolas Sep 13 '12 at 1:54