Showing that the radius of convergence for the complex powerseries $P(x)=\sum_{k=0}^{\infty}a_kx^k,P'(x)=\sum_{k=0}^{\infty}a_{k+n}x^{k}$ is equal

Question

$P'$ is not the derrivative

$n$ is fixed and $\in\mathbb{N}_0$

I wrote down the definitions and tried a proof by contradiction but I am stuck:

Radius of convergence can be calculated using the root-criterion for series in General.

$$R=\frac{1}{\limsup|\sqrt[k]a_k|},R'=\frac{1}{\limsup|\sqrt[k]a_{k+n}|}$$

$R$ is the radius of convergence for the powerseries $P(x)$ and $R'$ is the radius of convergence for the powerseries $P'(x)$.

Assume that $R \neq R'$

And without loss of generalitys $R'<R$

Then ther exists a $x_0\in\mathbb{C}$ such that $P(x_0)=L\in\mathbb{C}$ and $P'(x_0)$ diverges.

My idea is that to show that $P(x_0)=L\in\mathbb{C}\Rightarrow \exists_{L'\in \mathbb{C}} P'(x_0)=L'\leq L$. Which would then invoke a contradiction.

I wrote down the Definition of Limit which is

$$\forall_{\epsilon > 0}\exists_{k_0\in\mathbb{N}}\forall_{k\geq k_0}|\sum_{k_0=k}^{\infty}a_kx_0^{k}-L|<\epsilon$$

I now realize that this was not necessary because my goal was to find a convergent majorant for $P'(x_0)$. But Maybe someone of you can make use of it.

In the next step I tried to verify that

$$a_kx_0^{k}\geq a_kx_0^{k-n}(*)$$

Because then I could say

$$\sum_{k_0=k}^{\infty}a_kx_0^{k-n}\leq\sum_{k_0=k}^{\infty}a_kx_0^{k}$$

and would be done.

However the inequlatiy $(*)$ only holds if $x_0\geq 1$

Eddit: New Approach

Showing that if $|r|<R$

Then $P'(r)= \sum_{k=0} ^{\infty}a_k\frac{a_{k+n}}{a_k}r^k$ also converges

Answer 1

It's clear that both series converge for $z=0$. Now, suppose $|z|>0$. Let $P_m(z)$ and $P_m^{'}(z)$ be the partial sums of $P(z)$ and $P^{'}(z)$, respectively. Then, we have $$P^{'}_m(z) = \displaystyle\sum_{k=0}^{m} a_{k+n} z^{k}$$ $$P^{'}_m(z) = \dfrac{1}{z^n} \displaystyle\sum_{k=0}^{m} a_{k+n} z^{k+n}$$ $$P^{'}_m(z) = \dfrac{1}{z^n}\left( \displaystyle\sum_{k=0}^{m+n} a_{k} z^{k}-\displaystyle\sum_{k=0}^{n-1} a_{k} z^{k} \right)$$ $$P^{'}_m(z) = \dfrac{1}{z^n}\left(P_{m+n}(z)-\displaystyle\sum_{k=0}^{n-1} a_{k} z^{k} \right)$$ I'll use your notation for the radii of convergence. If we have that $R^{'} < |z| < R$ and take $m \rightarrow \infty$, we would have that the $RHS$ of the above equation converges and the $LHS$ diverges, a contradiction. We get a similar contradiction if we assume $R < |z| < R^{'}$. Thus, we conclude that $R = R{'}$.

Answer 2

If you know the formula for the radius of convergence, then the equality of the radii of convergence of the two series is straight-forward. We convenience, we work with the sequence $(\log(|a_k|)/k)$, the logarithm of $\sqrt[k]{|a_k|}$.

If we take a subsequence $(a_{k_i})_{i=1}^\infty$ for which $\lim_{i\to \infty} \frac{1}{k_i}\log(|a_{k_i}|)=\ell$ say, then we have

$$ \frac{\log(|a_{k_i}|)}{k_i-n} = \frac{k_i-n}{k_i}\left(\frac{\log(|a_{k_i}|)}{k_i}\right) \to 1.\ell $$ as $i\to \infty$. It follows that the sequences $(\frac{\log(|a_k|)}{k})_{k \geq 0}$ and $(\frac{\log(|a_{k}|)}{k-n})_{k\geq n}$ are such that their convergent subsequences have the same limits, and hence their $\limsup$s coincide.