About the limit $\lim_{x\to 1^-} (1-x)\sum_{n\ge 0} a_n x^n=L$

Question

If $(a_n)_{n\ge 0}$ is a sequence of complex numbers and $\lim_{n\to \infty} a_n=L\neq 0$ then $$\lim_{x\to 1^-} (1-x)\sum_{n\ge 0} a_n x^n=L$$

EDIT: The initial question had a typo from the place I'd seen it. Thank you guys for correction and help

Answer 1

If $\lim_na_n=L$ then $\lim_{x\to1^-}(1-x)\sum_{n=0}^\infty a_nx^n=L$.

Hint: The formula for the sum of a geometric series shows that $$(1-x)\sum a_nx^n=L+(1-x)\sum (a_n-L)x^n,$$so we need to show that $$\lim_{x\to1^-}(1-x)\sum(a_n-L)x^n=0.$$Let $\epsilon>0$. Choose $N$ so $|a_n-L|<\epsilon$ for all $n\ge N$. Hence $$\left|(1-x)\sum_{n=N}^\infty(a_n-L)x^n\right|<\epsilon$$for every $x\in(0,1)$. Show that $|(1-x)\sum_0^{N-1}(a_n-L)x^n|<\epsilon$ if $x$ is close enough to $1$.

Answer 2

This isn't true. Take $a_n = L$. Then you're claiming that $\lim_{x\to 1^-} \sum_{n\ge 0} L x^n=1-L$.