if $ S(x)=\sum\limits_{n=0}^{\infty}a_{n}x^n,|x|<R$, $S_k(R)$ bounded,prove or disprove $\lim\limits_{x\to R}S(x)$ exist?

Question

let $$S(x)=\sum_{n=0}^{\infty}a_{n}x^n,a_{n}>0,|x|<R$$ (or mean that powr series have radius of convergence R.)

and let $$S_{n}=\sum_{k=0}^{n}a_{k}x^k=a_{0}+a_{1}x+a_{2}x^2+\cdots+a_{n}x^n$$ if $S_{n}(R)$ is bounded,mean that there exsit constant $M>0$,such $$|S_{n}(R)|<M$$

prove or disprove $$\lim_{x\to R}S(x)$$ is exsit.?

I guess this is right,and I can't prove it.Thank you.

By the way: I have add this condition $a_{n}>0$.then I can't

Answer 1

Probably you know that if $|S_n(R)|<M$ and $a_n>0$, the series $\sum_{n=0}^\infty a_nR^n$ is convergent. If $R=1$, This is Theorem 8.2 of Baby Rudin and for arbitrary $R$ you can see following Theorem that you only need to define $g(x)=S(Rx)-S(R)$.

Complex Version of Abel's Limit Theorem. Let $f(z)=\sum_{n=0}^\infty a_nz^n$ have radius of convergence $R$ and let $z_0\in\mathbb C$ with $|z_0|=R$ where the series $f(z_0)=\sum_{n=0}^\infty a_nz_0^n$ converges. Then $$\lim_{z\to z_0}f(z)=f(z_0)$$ where the limit on the left is taken in such a way that $|z|<R$ and that the ratio $\frac{|z_0-z|}{R-|z|}$ remains bounded by a fixed constant.

Some part of proof that you need. By considering the limit of the function $$\color{red}{g(z)=f(z_0z)-f(z_0)}$$ as $z\to1$ in such a way that $|z|<1$ and that the ratio $\frac{|1-z|}{1-|z|}$ remains bounded by a fixed constant, we may henceforth assume that $z_0=1$ and that $f(z_0)=0$.

With these assumptions, if we put $s_n=a_0+a_1+...+a_n$, then $0=f(1)=\sum_{n=0}^\infty a_n=\lim s_n$. Now observe that $a_n=s_n-s_{n-1}$, so $$\sum_{k=0}^na_kz^k=(1-z)\sum_{k=0}^ns_kz^k+s_nz^n$$ Since $s_n\to0$ and $|z|<1$ it follows that $s_nz^n\to0$. Therefore, Taking $n\to\infty$, we obtain $$f(z)=\sum_{n=0}^\infty a_nz^n=(1-z)\sum_{n=0}^\infty s_nz^n.$$

Answer 2

Let $S_n(x)=\sum\limits_{k=0}^n a_kx^k$ and $L=\sum\limits_{k=0}^\infty a_kR^k$. From $a_k\ge0$, for every $n$ and $0<x<R$ we have $$ S_n(x) \le S(x) \le L. $$ From fixed $n$ and $x\to R-0$ you get $$ S_n(R) =\lim_{x\to R-0} S_n(x) \le \liminf_{x\to R-0} S(x) \le \limsup_{x\to R-0} S(x) \le L. $$ Then, from $n\to\infty$, $$ L=\lim_{N\to\infty} S_n(R) \le \liminf_{x\to R-0} S(x)\le \limsup_{x\to R-0} S(x) \le L. $$ Therefore $\liminf\limits_{N\to\infty} S(R) = \limsup\limits_{x\to R-0} S(x) =L$, so $\lim\limits_{x\to R-0} S(x)$ exists and $\lim\limits_{x\to R-0} S(x)=L=\sum\limits_{k=0}^\infty a_kR^k$.

Answer 3

You can also refer to Abel's continuity theorem. The function $S(x)$ is continuous in its convergence set; in particular if the series converges at $R$ then $S(x)$ is continuous at $R-0$.