Attention: $|z|<R$ where $R$ is the radius of convergence
Show that if $\sum_{k=0}^{\infty}a_kz^k$ converges for $z\in\mathbb{C}$ then $\sum_{k=0}^{\infty}a_{k+n}z^k$ also converges $n\in\mathbb{N}$
I have tried to solve this Problem for three Hours now here are my results so far but Nothing seems to work out:
I know since $|z|<R$ that the series converges absolutely.
A necessary condition for the convergence of a series is that the secuence of ist summands must be convergin to 0.
Therefore:
$\lim_{k\rightarrow \infty} |a_kz^k|=0$
Because the elements of the sequence are all positive the sequence must decrease at some Point.
That is without loss of generality that
$$|a_kz^k|\geq|a_{k+1}z^{k+1}|\geq|a_{k+2}z^{k+2}| \geq …\geq |a_{k+n}z^{k+n}|$$
Implying $\frac{|a_{k+n}z^{k+n}|}{|a_kz^k|}\leq 1\iff \frac{a_{k+n}}{a_k}\leq \frac{1}{z^n}$
Therefore
$\sum_{k=0}^{\infty}a_{k+n}z^k= \sum_{k=0}^{\infty} a_k\frac{a_k+n}{a_k}z^k\leq \sum_{k=0}^{\infty} |a_k\frac{a_k+n}{a_k}z^k|\leq \sum_{k=0}^{\infty}|a_k\frac{1}{z^n}z^k|$
And here is the Problem If I now say that
$\sum_{k=0}^{\infty}|a_k\frac{1}{z^n}z^k|\leq \sum_{k=0}^{\infty}|a_kz^k|$
Then I am assuming that $|z^{k-n}|\leq |z^k|\iff 1<|z^n|$
Therefore this does not work generally.
Please help me.
If $\sum_{k=0}^{\infty}a_kz^k$ converges absolutely so does $\sum_{k=0}^{\infty}a_{k+n}z^{k+n}$. Thus also $\sum_{k=0}^{\infty}a_{k+n}z^k$ for $z\not=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $\sum_{k=0}^{\infty}a_{k+n}z^k$ is at least that of $\sum_{k=0}^{\infty}a_kz^k$. If, on the other hand, $\sum_{k=0}^{\infty}a_{k+n}z^k$ converges absolutely so does $\sum_{k=0}^{\infty}a_{k+n}z^{k+n}$ and therefore also $\sum_{k=0}^{\infty}a_kz^k=\sum_{k=0}^{n-1 }a_kz^k+\sum_{k=0}^{\infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $\sum_{k=0}^{\infty}a_kz^k$ is at least that of $\sum_{k=0}^{\infty}a_{k+n}z^{k+n}$.
Thus they are equal.