Hi everyone ! I need a help with this stuff
Show that a power series c(sub n)x^n has the same radius of convergence as c(sub n+m)x^n, for any positive integer m.
Thanks.
Suppose that sum series $\displaystyle \sum{c_nx^n}$ converges $\displaystyle \forall{x}\in(-\delta,\delta)$. This implies that $\displaystyle \lim_{n\to\infty}\sqrt[n]{|c_nx^n|}<1~\forall{x}\in(-\delta,\delta)$
Now we are looking at $\displaystyle \sum{c_{n+m\in\mathbb{N}}x^n}$. So using the ratio test we need to find the value of x such that $\displaystyle \lim_{n\to\infty}\sqrt[n]{|c_{n+m}x^n|}<1$. Now letting $\displaystyle n+m=z$ we have $\displaystyle \lim_{z\to\infty}\sqrt[z-m]{c_zx^{z-m}}=|x|\lim_{z\to\infty}\sqrt[z-m]{c_z}\sim|x|\lim_{x\to\infty}\sqrt[z]{c_z}<1~\forall{x}\in(-\delta,\delta)\quad\blacksquare$
both series are convergent at x = 0. for $\displaystyle x \neq 0$ let $\displaystyle p(x)=c_0 + c_1x + \cdots + c_{m-1}x^{m-1}.$ now the claim trvially follows from this identity: $\displaystyle \sum_{n=0}^{\infty}c_nx^n = x^m\sum_{n=0}^{\infty}c_{n+m}x^n +p(x). \ \ \Box$