Show that two power series have the same radius of convergence

Question

Show that the power series $\sum_{n=0}^{\infty}{c_{n}x^{n}}$ has the same radius of convergence as $\sum_{n=0}^{\infty}{c_{n+m}x^{n}}$ for any positive integer $m$.

Should I use the Ratio Test? Or do I use induction?

Answer 1

The radius of convergence $R$ can be characterized as follows:

1. If $|x|<R$, then the series $\sum\limits_{n\ge0}c_nx^n$ converges
2. If $|x|>R$, then the series $\sum\limits_{n\ge0}c_nx^n$ does not converge

Let's put aside the cases $R=0$ and $R=\infty$, for the moment.

There is no problem with convergence at $0$, so we can assume $x\ne0$.

Suppose $|x|<R$. Then the series $\sum\limits_{n\ge0}c_nx^n$ converges, so also $$ \sum_{n\ge1}c_nx^n $$ does and, by multiplying by $1/x$, also $$ \frac{1}{x}\sum_{n\ge1}c_nx^n=\sum_{n\ge1}c_nx^{n-1}= \sum_{n\ge0}c_{n+1}x^n $$ converges.

Conversely, if $\sum\limits_{n\ge0}c_{n+1}x^n$ converges, then also $$ x\sum_{n\ge0}c_{n+1}x^n=\sum_{n\ge0}c_{n+1}x^{n+1}=\sum_{n\ge1}c_{n}x^n $$ converges and so does $\sum\limits_{n\ge0}c_nx^n$.

Now you can start induction.

For the case $R=\infty$ it's the same, because it means that the series $\sum\limits_{n\ge0}c_nx^n$ converges for all $x$.

Do separately the case $R=0$.

Answer 2

First of all, note that one has always: $$\frac{1}{x}\left(\sum_{n=0}^Nc_nx^{n+1}\right)=\frac{1}{x}\left(x\sum_{n=0}^Nc_nx^n\right)=\sum_{n=0}^Nc_nx^n\quad(x\neq0)$$ So all expressions give rise to identical series!

Now, let's show the equivalence: $$\sum_{n=0}^\infty c_nx^n\in\mathbb{C}\iff x\sum_{n=0}^\infty c_nx^n\in\mathbb{C}$$ On the one hand one has for nonzero values: $$\sum_{n=0}^\infty c_nx^n\in\mathbb{C}\implies x\sum_{n=0}^\infty c_nx^n\in\mathbb{C}$$ On the other hand one has for nonzero values: $$x\sum_{n=0}^\infty c_nx^n\in\mathbb{C}\implies\frac{1}{x}\left(x\sum_{n=0}^\infty c_nx^n\right)\in\mathbb{C}$$ And the trivial case can be checked separately: $$\sum_{n=0}^\infty c_n0^n\in\mathbb{C}\iff0\sum_{n=0}^\infty c_n0^n\in\mathbb{C}$$