Radius of convergence of power series and its formal derivative Suppose that power series $\sum\limits_{n=0}^{\infty} c_n(z-z_0)^n$ has radius of convergence $R$, i.e., it converges absolutely on $|z-z_0|<R$ and diverges on $|z-z_0|>R$. Consider the series obtained after formal differentiation $\sum\limits_{n=0}^{\infty} (n+1)c_{n+1}(z-z_0)^n$. I want to show that it has the same radius of convergence $R$.
By Cauchy-Hadamard theorem it is enough to show that $$\limsup \limits_{n\to \infty}\sqrt[n]{|c_n|}=\limsup \limits_{n\to \infty}\sqrt[n]{(n+1)|c_{n+1}|}.$$
But since $\limsup \limits_{n\to \infty}\sqrt[n]{(n+1)|c_{n+1}|}=\limsup \limits_{n\to \infty}\sqrt[n]{|c_{n+1}|}$ so we need to show that $$\limsup \limits_{n\to \infty}\sqrt[n]{|c_n|}=\limsup \limits_{n\to \infty}\sqrt[n]{|c_{n+1}|}.$$
I know that this question has been asked couple times but I asked in those posts and no one answered me since topics are outdated.
I would be very thankful for clear explanation!
 A: I'm gonna give a more conceptual proof.
The series $\sum_{n=0}^\infty (n+1)c_{n+1}z^n$ and $\sum_{n=0}^\infty (n+1)c_{n+1}z^{n+1}$ have the same radius of convergence because the latter is the former multiplied by $z$.
The series $\sum_{n=0}^\infty (n+1)c_{n+1}z^{n+1}$ and $\sum_{n=0}^\infty c_nz^n$ have the same radius of convergence by a straightforward calculation.
We conclude the series $\sum_{n=0}^\infty (n+1)c_{n+1}z^n$ and $\sum_{n=0}^\infty c_nz^n$ have the same radius of convergence.
No need to do any weird computation here ;)
A: Let$$a_n=\sqrt[n]{|c_n|}\quad\text{and}\quad b_n=\left(a_{n+1}\right)^{(n+1)/n}=\left(\sqrt[n+1]{|c_{n+1}|}\right)^{(n+1)/n}=\sqrt[n]{|c_{n+1}|}.$$
If we prove that $\limsup_{n\to\infty}a_n=\limsup_{n\to\infty}b_n$, then the problem is solved.
Note that this defines $b_n$ using $a_n$, but the process can be reversed (that is, we can define $a_n$ using $b_n$), since$$n>1\implies a_n=(b_{n-1})^{n/(n-1)}.\tag1$$
Let $S=\limsup_{n\to\infty}a_n$. There is a subsequence $(a_{n_k})_{k\in\Bbb N}$ of $(a_n)_{n\in\Bbb N}$ such that$$\lim_{k\to\infty}a_{n_k}=S.\tag2$$But then$$\lim_{k\to\infty}b_{n_k-1}=\lim_{n\to\infty}(a_{n_k})^{n_k/(n_k-1)}=S,$$because we have $(2)$ and also because $\lim_{k\to\infty}\frac{n_k}{n_k-1}=1$. So, $\limsup_{n\to\infty}b_n\geqslant S$. But if $\limsup_{n\to\infty}b_n=S^\star>S$, then the same argument, together with $(1)$, shows that$$\limsup_{n\to\infty}a_n\geqslant S^\star>S=\limsup_{n\to\infty}a_n,$$which is impossible.
A: Let's first consider the case when $R\in (0,\infty)$.
Let $\frac 1R=\limsup |c_n|^\frac 1n$. Choose any $r\gt \frac 1R$.
It follows that for all large $n$, $|c_n|^\frac 1n\lt r$, which gives $|c_{n+1}|^{\frac 1{n+1}}\lt r$ for all large $n$.
Simplifying gives: $$|c_{n+1}|^\frac 1n\lt r^{\frac {n+1}n}=r. r^\frac 1n \text{ for all large $n$ }\implies \limsup |c_{n+1}|^\frac 1n\le r\tag 1$$

Note that $x:=\limsup |c_{n+1}|^\frac 1n$ can't be smaller than $\frac 1R$. For if $x\lt \frac 1R$ then choose $y\in (x,\frac 1R)$ and hence by limit superior definition it follows that $|c_{n+1}|^\frac 1n\lt y$ for all large $n$, which is same as saying $|c_{n+1}|^\frac 1{n+1}\lt y^{\frac n{n+1}}$ for all large $n$. It follows that $$\limsup |c_{n+1}|^{\frac 1{n+1}}\le y\implies \frac 1 R\le y$$ which is a contradiction.

The following must hold:$$x\ge \frac 1R\tag 2$$
It follows by $(1)$ and $(2)$ that $\limsup |c_n|^\frac 1n=x$.
Particular cases when $R=0$ (with the notion that $1/0:=\infty$) and $R=\infty$ (with the notion that $\frac 1\infty:=0$) can be handled separately.
A: Maybe this will be helpful: if $x_n \to a \gt 0$, then $x_n^{\frac{n+1}{n}}=x_n^{1+\frac{1}{n}}\to a$, as $x_n^{\frac{1}{n}}\to 1$.
Addition.
Let's consider $a_n=x_n^{\frac{1}{n}} = x_n^{\frac{1}{n+1}\cdot \frac{n+1}{n}} =b_n^{\frac{n+1}{n}}$, so $b_n=x_n^{\frac{1}{n+1}}$. As is written above and you agreed in comment, if $b_n \to b \gt 0$ have some limit, then same limit will have $a_n$. So, we can say same for positive limit points of $b_n$, which gives, that $a_n$ have same positive limit points with $b_n$.
