Is $\limsup |C_n|^{\frac{1}{n}} = \limsup |C_{n+j}|^{\frac{1}{n}}? $ where $j\geq0 $

Question

$C_n$ 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?

Answer 1

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 $\limsup a_n^{b_n} = \limsup a_n.$