Show $\sum_{n=0}^{\infty}a_nx^n$ has same radius of convergence as its derivative Prove that if $\sum_{n=0}^{\infty}na_nx^{n-1}$ converges for $|x| \lt r$, then $\sum_{n=0}^{\infty}a_nx^n$ also converges for $|x| \lt r$, i.e. have the same radius of convergence. 
I tried to apply comparison test, but only get absolute convergence, i.e. I know how to prove if $\sum_{n=0}^{\infty}|na_nx^{n-1}|$ converge then $\sum_{n=0}^{\infty}|a_nx^n|$ also converges. But I'm stuck on showing the above property. Could someone give a proof please? Thanks.
 A: A power series $\sum a_nx^n$ converges absolutely if $|x| < R$ and diverges if $|x| > R$, where the radius of convergence can be found as 
$$R = \left(\limsup_{n \to \infty}|a_n|^{1/n}\right)^{-1}.$$
Note that
$$\limsup_{n \to \infty}|na_n|^{1/n} \leqslant \limsup_{n \to \infty}|n|^{1/n} \limsup_{n \to \infty}|a_n|^{1/n}= 1 \cdot \limsup_{n \to \infty}|a_n|^{1/n}, $$
and, since $|a_n| \leqslant |na_n|,$
$$\limsup_{n \to \infty}|a_n|^{1/n} \leqslant \limsup_{n \to \infty}|na_n|^{1/n}.$$
Hence,
$$\limsup_{n \to \infty}|na_n|^{1/n} = \limsup_{n \to \infty}|a_n|^{1/n}.$$
A: You want to show that,
if
$R 
= \left(\limsup_{n \to \infty}|na_n|^{1/n}\right)^{-1}
$
then
$R 
= \left(\limsup_{n \to \infty}|a_n|^{1/n}\right)^{-1}
$.
This follows from
$\lim_{n \to \infty}
n^{1/n}
= 1
$.
A: If $\sum n a_n x^n$ converges for $|x|<r$, then for any $0<r_1<r$ there holds $\lim_{n\to \infty}na_nr_1^n=0$. In particular, for all large enough $n$ you get $|na_nr_1^n|<1$. This means that $|a_n|< 1/(nr_1^n)<1/r_1^n$ for $n$ large enough. 
Now if you want to show for a given $x,|x|<r$ that $\sum a_n x^n$ converges, then pick $r_1 with $|x|

The argument is a bit more subtle in the other direction (that the derivative has
radius of convergence at least as big as the original), since the inequality would be $|a_n|<n/r_1^n$. So you would end up comparing with
$\sum nq^n$ with $0<q<1$. While not a geometric series, it also converges, for example by a ratio test.
