Radius of convergence of a sum of power series

You can show that it is the minimum of the 2 convergence radius, if the are different. This one is easy as for $x\leq \min\{\rho_1,\rho_2\}$ the convergence of the sums is absolute. So it is just simple arithmetic $$\sum_{n=1}^\infty (a_k +b_k ) x^k = \sum_{n=1}^\infty a_k x^k + \sum_{n=1}^\infty b_k x^k$$

When they are the same, you only can say that it is greater equal than the convergence radius. Taking for example $a_k=-1$ and $b_k=1$ the convergence radius of $$ \sum_{k=1}^\infty (a_k+b_k) x^k $$ is infinity.

To see that if the radius are different we really only have the minimum and not more as the convergence radius we make this here:

Let us say the convergence radius of $\sum_{k=1}^\infty a_k x^k $ is $\rho_1$ and the one of $\sum_{k=1}^\infty b_k x^k$ is $\rho_2$ we say now $\rho_1 < \rho_2$,

If the convergence radius would be greater than $\rho_1$ than the following equation must be true: $$\sum_{k=1}^\infty (a_k +b_k) x^k -\sum_{k=1}^\infty b_k x^k = \sum_{k=1}^\infty a_k x^k. $$ As on the left hand side we have the difference of two convergent series the right hand side must be convergent too, but it is only convergent if $x\leq \rho_1$.

Since

$\displaystyle\sum_{n=1}^{\infty} (a_n +b_n)x^{n}=\displaystyle\sum_{n=1}^{\infty} a_n x^{n}+\displaystyle\sum_{n=1}^{\infty} b_n x^{n} $ it is clear that the left hand side converges for values of $x$ exactly when both of the series on the RHS converge.

Since this means that the LHS will converge on the intersection of the discs of convergence on the RHS, you know the radius of convergence for the LHS is the same as the smaller radius on the RHS.

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I only have confidence in this answer for absolutely convergent series. Perhaps its necessary after all to resort to a test for convergence, rather than relying on this naive arithmetic.