Proof, that multiplying a power series with $\frac{1}{x^k} doesn't change its radius of convergence.

Question

Let $f: x\mapsto \sum_{i = 0}^{\infty} a_nx^n$ be a power series with convergence radius $R_f$ and $k\in\mathbb{N}$, where $a_i = 0,\ \forall i=0,...,k-1$. Multiplying the series with $\frac{1}{x^k}$ can be converted to $$\frac{1}{x^k} \sum_{i=0}^{\infty}a_nx^n = \sum_{i=-k}^{\infty}a_{n+k}x^n = \sum_{i=-k}^{-1}a_{n+k}x^n + \sum_{i=0}^{\infty}a_{n+k}x^n = \sum_{i=0}^{\infty}a_{n+k}x^n$$ Let $g: x\mapsto \sum_{i=0}^{\infty}a_{n+k}x^n$ with radius of convergernce $R_g = \frac{1}{\limsup_{n\to\infty}\sqrt[^{n}]{a_{n+k}}}$. One can use the following estimation: $$ {a_{n+k}}^{\frac{1}{n}} \ge {a_{n+k}}^{\frac{1}{n+k}} \quad\Leftrightarrow \quad \frac{1}{n} \ge \frac{1}{n+k}\quad \Leftrightarrow \quad1 + \frac{k}{n} \ge 1 \quad\Rightarrow \quad \sqrt[^{n}]{a_{n+k}} \ge \sqrt[^{n+k}]{a_{n+k}} , \; \forall n\in\mathbb{N} $$ As $\limsup_{n\to\infty}\sqrt[^{n+k}]{a_{n+k}} = \limsup_{n\to\infty}\sqrt[^{n}]{a_{n}}$, the inequations $\limsup_{n\to\infty}\sqrt[^{n}]{a_{n+k}} \ge \limsup_{n\to\infty}\sqrt[^{n}]{a_{n}}$ and $R_g \le R_f$ follow. Case by case analysis of $R_f$:

• $R_f = 0$ : $R_g = R_f = 0$
• $1 \le R_f < \infty$ : Given that $\limsup_{n\to\infty} a_n = 0$ we now have the estimate $\limsup_{n\to\infty}\sqrt[^{n}]{a_{n}} \ge \limsup_{n\to\infty}\sqrt[^{n}]{a_{n+k}}$, combined $R_f \ge R_g \ge R_f$ and therefore $R_g = R_f$

I'm not sure how to approach the cases $R_f = \infty$ and $0 < R_f < 1$ - or whether my previous conclusions are even correct for that matter - so I'd be grateful about some help.

Answer 1

There is a much simpler way to go about this. We just need some simple results for sequences. Let $a_n$ and $b_n$ be complex valued sequences.

1. If $a_n\to a$ and $b_n\to b$, then $a_nb_n\to ab$. In particular, the product of convergent sequences is convergent.
2. If $a_n\to a\neq0$ and $b_n$ does not converge, then neither does $a_nb_n$.

You can look these up or try to prove them yourself. Using these, we can see:

1. Let $a_n(x)=x^{-k}$ and $b_n(x)$ a power series of the type you mentioned above, where $\vert x\vert$ is less than the radius of convergence of $b_n(x)$. Then both $a_n$ and $b_n$ converge, and the sequence $a_nb_n$ is the "shifted" power series from above, and it converges.
2. Now let $\vert x\vert$ be greater than the radius of convergence. In particular, $x^{-k}\neq0$. Then $b_n(x)$ diverges, and $a_n(x)$ converges to something other than 0. So the product $a_nb_n$ (which is still the shifted sequence from above) diverges.

Together, these imply that the radius of convergence of $b_n(x)$ and $a_n(x)b_n(x)$ are the same.