Prove that series $ \sum^{+\infty}_{n=0}a_n(x-x_0)^n $ and $ \sum^{+\infty}_{n=0}(n+1)a_{n+1}(x-x_0)^n $ have the same radius of convergence. I want to prove that these two power series
$$
\sum^{+\infty}_{n=0}a_n(x-x_0)^n
$$
and
$$
\sum^{+\infty}_{n=0}(n+1)a_{n+1}(x-x_0)^n
$$
have the same radius of convergence.

What I've done so far is:


*

*I replace $(x-x_0)$ with $y$ to make calculations easier.

*In order to find the radius I use the Cauchy's criterion (Root test):
1:
$$
\lim_{n \to \infty}\sqrt[n]{a_ny^n} < 1\\
y\lim_{n \to \infty}\sqrt[n]{a_n} < 1\ \ \ (*)\\
$$
2:
$$
\lim_{n \to \infty}(\sqrt[n]{(n+1)a_{n+1}y^n}) < 1\\
y\lim_{n \to \infty}(\sqrt[n]{n+1})\lim_{n \to \infty}(\sqrt[n]{a_{n+1}}) < 1\\
y\lim_{n \to \infty}(\sqrt[n]{a_{n+1}}) < 1 \ \ \ (**)
$$

*I haven't found the radius, but if I take the limit of $(*)$ and $(**)$ as $n \to \infty$ then $(*)$ and $(**)$ should be equal.

Is my reasoning correct? Can we actually get a formula for the radius?
 A: Yes, you can apply the formula of the radius of convergence. Your solution is essentially correct. Here is a more direct way to present it.
Given the series 
$$
S_1(x)= \sum^{+\infty}_{n=0}a_n(x-x_0)^n
$$
its radius of convergence $\rho_1$ is 
$$\frac{1}{\rho_1} = \limsup_{n \to \infty}  \sqrt[n]{\vert a_n  \vert}$$ 
Given the series 
$$
S_2(x)= \sum^{+\infty}_{n=0}(n+1)a_{n+1}(x-x_0)^n
$$
then, its radius of convergence $\rho_2$ is
$$\frac{1}{\rho_2} = \limsup_{n \to \infty}  \sqrt[n]{(n+1)\vert a_{n+1}  \vert}$$
Now, since we know $\lim_{n \to \infty}\sqrt[n]{n+1}=1$, we have 
$$\frac{1}{\rho_2} = \limsup_{n \to \infty}  \sqrt[n]{(n+1)\vert a_{n+1}  \vert}=(\lim_{n \to \infty}\sqrt[n]{n+1})\limsup_{n \to \infty}  \sqrt[n]{\vert a_{n+1}  \vert}=\limsup_{n \to \infty}  \sqrt[n]{\vert a_n  \vert}=\frac{1}{\rho_1}$$
So we have $\rho_1=\rho_2$.
A: Regarding the last question on whether or not one can get a formula, the Cauchy--Hadamard theorem says that the radius of convergence $R$ of $\sum_{n=0}^\infty a_n (x-x_0)^n$ satisfies $$\frac{1}{R} = \limsup_{n \to \infty} \lvert a_n \rvert^{1/n},$$
where the possibilities $0$ and $\infty$ are treated in the natural way.
A: Your solution seems to be reasonable, however I believe that there is an easier way to prove the statement.

Note that the second series is, actually, a derivative of the first. 
Indeed, denote 
$F(x) = \sum^{+\infty}_{n=0}a_n(x-x_0)^n$ and 
$G(x) = \sum^{+\infty}_{n=0}(n+1) a_{n+1}(x-x_0)^n$. 
Take derivative of $F$:
$$
F'(x) = \left(\sum^{+\infty}_{n=0}a_n(x-x_0)^n\right)' = 
\left(a_0 + \sum^{+\infty}_{n=1}a_n(x-x_0)^n\right)' = 
\sum^{+\infty}_{n=1}na_n (x-x_0)^{n-1}
$$
Let us shift summation index by $-1$ by making change of variables $m = n-1$. 
Then we get 
$$
F'(x) =  \sum^{+\infty}_{n=1}na_n (x-x_0)^{n-1}
= \sum^{+\infty}_{m=0}(m+1)a_{m+1} (x-x_0)^{m}
$$
Without Loss of generality, we can re-lable $m$ back to $n$, thus getting 
$$
F'(x) = \sum^{+\infty}_{n=0}(n+1)a_{n+1} (x-x_0)^{n} = G(x)
$$
As we know, differentiation and integration does not change radius of convergence of power series. 
Thus,  the radius of convergence of $F$ and $G$ is the same.
Q.E.D.
