let $ f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n} $ be with positive convergence radius $ R>0 $ and assume that exists sequence $ y_n \in(0,R) $ decreasing monotonic such that $ y_{n}\to0 $ and for any $ n\in \mathbb{N} $ $ f\left(y_{n}\right)=0 $. we need to prove that $ a_n=0 $ for any $ n\in \mathbb{N} $ what i've done:
prove by induction :
for $n=0 $ we have $ f\left(0\right)=\sum_{n=0}^{\infty}a_{n}0^{n}=a_{0} $
also the series is cointinious, therefore $ \lim_{x\to0}f\left(x\right)=f\left(0\right)=a_{0} $
so using Heine's theorem:
$ \lim_{x\to0}f\left(x\right)=\lim_{n\to\infty}f\left(y_{n}\right)=a_0 $
and since for any $ n\in \mathbb{N} $ $f(y_n)=0 $ we can conclude that $a_0=0 $
Now assume the statement is true for any $n<m$ for some $m\in \mathbb{N} $
So we have:
$ 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) $
let $ k\in\mathbb{N} $. from the given condintions $ f\left(y_{k}\right)=0 $ and $ y_{k}\neq0 $ so
$ f\left(y_{k}\right)=\left(y_{k}\right)^{m}\left(\sum_{n=m}^{\infty}a_{n}\left(y_{k}\right)^{n-m}\right)=0 $
Thus, we can colclude that $ \sum_{n=m}^{\infty}a_{n}\left(y_{k}\right)^{n-m}=0 $ for any $ k\in \mathbb{N} $
Name this function $ g\left(x\right)=\sum_{n=m}^{\infty}a_{n}x^{n-m} $
and we know that the function is continious, therefore
$ \lim_{x\to0}g\left(x\right)=g\left(0\right)=a_{m} $
and using Heine's theorem once again gives us:
$ \lim_{x\to0}g\left(x\right)=\lim_{n\to\infty}g\left(y_{n}\right)=a_{m}=0 $
therefore $a_m$ =0
That proves the theorem, but I didnt use the fact that $y_n $ is decreasing. I cant see where I was wrong. any thoughts will be helpful.