I'm trying to explain why these two series $$\sum_{n=1}^{\infty}nc_{n}(z-a)^{n}$$ and$$\sum_{n=1}^{\infty}nc_{n}(z-a)^{n-1}$$ have the same radius of convergence.
I know that $\sum_{n=1}^{\infty}nc_{n}(z-a)^{n}=(z-a)\sum_{n=1}^{\infty}nc_{n}(z-a)^{n-1}$ and knowing that I should know how to conclude that the set of points that both converge are the same, but I don't.
On the other hand, in most of the tests that I use I know that if some limit exist then the radius of convergence is this limit.
In those tests mentioned it doesn't require more than $c_n$ (which both series have in common) but if there is a radius of convergence I am not sure that it implies that the test should give it as a consequence, and because of that I don't know how to proceed. Any ideas? Thanks in advance!!
If $\lambda\in\mathbb C\setminus\{0\}$, then a series $\sum_{n=1}^\infty a_n$ converges if and only if the seres $\sum_{n=1}^\infty\lambda a_n$ converges. Therefore,$$z\neq a\implies\sum_{n=1}^\infty nc_n(z-a)^n\text{ converges if and only if }\sum_{n=1}^\infty nc_n(z-a)^{n-1}\text{ converges.}$$Besides, both series converge when $z=a$. So$$\left\{z\in\mathbb C\,\middle|\,\sum_{n=1}^\infty nc_n(z-a)^n\text{ converges}\right\}=\left\{z\in\mathbb C\,\middle|\,\sum_{n=1}^\infty nc_n(z-a)^{n-1}\text{ converges}\right\}$$and therefore both series have the same radius of convergence.