Does the radius of convergence for a power series change when you multiply it by a constant?

I have to use differentiation to find power series representations for certain functions. I also have to find the radius of convergence. For part a) of the problem, I have to do this for $f(x)= \frac {1}{(1+x^2)}$. I then have to use the answer for that problem to find the power series representation for $f(x)= \frac {1}{(1+x^3)}$. However, in order to do so, I have to modify my answer from part a) and multiply it by $-\frac{1}{2}$. This brings me to my question: does the radius of convergence for a power series change when you multiply it by a constant? What if it wasn't a power series but some other type of series such as a rational or algebraic one, would it change then?

• Regarding your last question, the notion of "radius of convergence" is not necessarily defined for non-power series. If $f(x)$ is defined by some arbitrary series, it isn't necessarily true that convergence for some particular $x$ implies convergence for all smaller $x$.
– user169852
Commented Mar 21, 2018 at 6:48

As long as the constant is not $0$, the radius of convergence is not changed. This is because if $c \ne 0$, a series $\sum_n c a_n$ converges if and only if $\sum_n a_n$ converges. By the way, when they converge $\sum_n c a_n = c \sum_n a_n$.