Let $\phi(t)=\sum_{k=1}^\infty a_k t^k$, $x=t^m \in \mathbb{C}$ for some fixed $m\in \mathbb{N}$ be a convergent power series. I guess that $a_0=0$. For $r=0,\ldots,m-1$ and $k=mq+r$, why are the power series

$$\tilde\phi_r(x) = \sum_{q=0}^\infty a_{mq+r}x^q$$


The issue is that I cannot deduce from the convergence of the first series (and a probably associated convergence radius $r$) anything that could provide the convergence of the second one.


The Cauchy-Hadamard formula for the radius of convergence does the trick. Suppose

$$\frac{1}{R_0} = \limsup_{k\to\infty} \lvert a_k\rvert^{1/k} < \infty$$

(so the original series converges somewhere). Then

$$\begin{align} \limsup_{q\to\infty} \lvert a_{qm+r}\rvert^{1/q} &= \limsup_{q\to\infty} \left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^{m+r/q}\\ &\leqslant \limsup_{q\to\infty}\left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^m \cdot \limsup_{q\to\infty} \left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^{r/q}\\ &\leqslant \frac{1}{R_0^m}\cdot 1, \end{align}$$

thus the radius of convergence of the slice is at least the $m$-th power of the radius of convergence of the full series.


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