Is $\limsup |C_n|^{\frac{1}{n}} = \limsup |C_{n+j}|^{\frac{1}{n}}? $ where $j\geq0 $ $C_n$ is a sequence of complex numbers.
I would like to prove that the radius of convergence of a differentiated power series is the same as the original series.
i.e. $\limsup |(n+j)(\cdots)(n+1)C_{n+j}|^{\frac{1}{n}} = \limsup |C_n|^{\frac{1}{n}}$
I know that $\limsup |(n+j)(\cdots)(n+1)C_{n+j}|^{\frac{1}{n}} = \limsup|C_{n+j}|^{\frac{1}{n}}$ since $\lim((n+j)(\cdots)(n+1))^{\frac{1}{n}}=1$
But I am stuck at the above question.
Any help?
 A: I'll assume $j=1$ (the case of arbitrary $j$ is analogous, and actually directly addressed by the proposition in the block below).
The question is then equivalent to asking whether, for a positive sequence $a_n$,
$$\limsup a_n^{1/(n-1)}=\limsup a_n^{1/n}.$$
By putting the left side as $(a_n^{1/n})^{(n/(n-1))}$, we have that the question will be  answered if we can prove the following statement:

If $b_n$ is a sequence which converges to $1$ and $a_n>0$, then
  $$\limsup a_n^{b_n} =\limsup a_n. $$

For that, we can proceed as follows: fix $\epsilon>0$.  For sufficiently large $n$, it holds that
$$a_n^{1-\epsilon}\leq a_n^{b_n} \leq a_n^{1+\epsilon}. $$
Therefore,
$$\limsup a_n^{1-\epsilon} \leq \limsup a_n^{b_n} \leq \limsup a_n^{1+\epsilon}, $$
which implies
$$(\limsup a_n)^{1-\epsilon} \leq \limsup a_n^{b_n} \leq (\limsup a_n)^{1+\epsilon}.$$
Since this holds for every $\epsilon>0$, we have
$$\limsup a_n\leq \limsup a_n^{b_n} \leq \limsup a_n,$$
which implies $\limsup a_n^{b_n} = \limsup a_n.$
