Consider the sequence $a_n = (-1)^nn$ and let $f(x)= \sum_{i=1}^\infty a_nx^n$
Show that there is a continuous function $g:(-1,1]\to R$ such that $g(x) = f(x)$ for any $x\in (-1,1)$
So, I understand that f(x) is continuous on $(-1,1)$ (since it is uniformly convergent there) but how do I extend this to $x=1$? I was also given a hint to consider $h(x)= \sum_{i=1}^\infty (-1)x^n$ and $xh'x$ which I know is equal to $f(x)$, but I still can't understand the $x=1$ case. Thank you!
For each $x\in(-1,1)$,$$f(x)=\sum_{n=1}^\infty(-1)^nnx^n=-\frac x{(1+x)^2}.$$So, simply take $g(x)=-\frac x{(1+x)^2}$, for each $x\in(-1,1]$.