let $ f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n} $ absolute convergent series with convergent radius $ R>0 $
Assume $ a_0,a_1,a_2,...a_{m-1}=0 $
so $ f\left(x\right)=\sum_{n=m}^{\infty}a_{n}x^{n}=x^{m}\left(\sum_{n=m}^{\infty}a_{n}x^{n-m}\right) $
I want to claim the the series $ \sum_{n=m}^{\infty}a_{n}x^{n-m} $
also convergent with the same radius. Is this way of proving it correct ? :
use the limit comparison test:
$ |\frac{a_{n}x^{n-m}}{a_{n}x^{n}}|=|x^{-m}|\underset{n\to\infty}{\longrightarrow}|x^{-m}| $
so for every $ x\in\left(0,R\right)\cup\left(-R,0\right) $ the series absolutely convergent by limit comparison test. for $ x=0 $ its trivial.
Is it correct or am i missing something? Thanks.
Yes, this works out. One could also note that since one can calculate the radius of convergence explicitely as the inverse of $\limsup_{n \rightarrow \infty} \vert a_n\vert^{1/n}$. Then your question comes down to asking whether a sequence and any of its tails converge to the same value, which is true.