How to show that $\sum\limits^\infty_{n=0}a_n(\frac{2x}{1+x})^n$ is continuous?

Question

$(a_n)$ is a bounded sequence and $x\in (-1,1)$. I have shown point wise convergence. I have tried using Weierstrass but failed, and the only other thing I can think of is using Abel's Theorem for power series, but I am unsure if it applies.

Thanks in advance!

Answer 1

This is a consequence of the fact that the composition of continuous functions is continuous. The function $f(x) = {2x \over 1 + x}$ is continuous on ${\mathbb R} - \{-1\}$, and the power series $\sum_n a_n y^n$ is continuous on $(-R,R)$ where $R$ is the radius of convergence of the power series.

So the composition $\sum_n a_n ({2x \over 1 + x})^n$ is continuous on $(a,b)$, where $(a,b) = f^{-1}(-R,R)$.

Note that $f^{-1}(-R,R)$ does not have to contain $(-1,1)$. For example, if $a_n = 1$ for all $n$ then $R = 1$ and therefore $(a,b) = (-{1 \over 3},1)$. So $\sum_n a_n ({2x \over 1 + x})^n$ is not even defined for $x < -{1 \over 3}$.

Answer 2

Let $f(x)=\dfrac{2x}{1+x}$. We prove that $\sum\limits^\infty_{n=0}a_nf^n$ is continuous on $(-\frac{1}{3},1)$.

First consider $[-\frac{r}{3},r],\:r<1$.

Since $$ f'=\dfrac{2}{(1+x)^2}>0\: $$ for $x\in[-\frac{r}{3},r],\:$ there is $$ -\dfrac{2r}{3-r}<f(x)<\dfrac{2r}{1+r} $$ Since either $$ \sum\limits^\infty_{n=0}|a_nf^n|=\sum\limits^\infty_{n=0}\left|a_n\left(\frac{2x}{1+x}\right)^n\right|<M\sum\limits^\infty_{n=0}\left(\frac{2r}{1+r}\right)^n=M\frac{1+r}{1-r} $$ Or $$ \sum\limits^\infty_{n=0}|a_nf^n|<M\sum\limits^\infty_{n=0}\left(\frac{2r}{3-r}\right)^n=M\frac{1-r/3}{1-r} $$ converge on $[-\frac{r}{3},r]$. By Weierstrass's M test, $\sum\limits^\infty_{n=0}a_nf^n$ is uniform convergent on $[-\frac{r}{3},r]$ and is continuous since $f^n$ is continuous.

Finally since for any $x\in(-\frac{1}{3},1)$, there is a $r,\:-\frac{r}{3}<x<r<1$. So we can prove $\sum\limits^\infty_{n=0}a_nf^n$ is continuous on $(-\frac{1}{3},1)$.