Show $\lim_{j\rightarrow\infty}\sum_{k=0}^\infty a_k z_j^k=\sum_{k=0}^\infty a_k(-1)^k$ Can someone help me with this proof, or guide me? I'm thinking I need to show that
$$
\left|
  \sum_{k=0}^\infty a_k z_j^k-\sum_{k=0}^\infty a_k(-1)^k
\right| <\epsilon \quad \forall j>M\in \mathbb{Z}$$ but I don't know how to continue after.
Let $\{a_k\}_{k=0}^\infty$ be a sequence of complex numbers such that $\sum_{k=0}^\infty a_k(-1)^k$ converges. If $\{z_k\}$ is a sequence of complex numbers such that $|z_j|<1$, $\frac{|1+z_j|}{1-|z_j|}\leq4$ and $\lim_{j\rightarrow\infty}z_j=-1$, show that
$$\lim_{j\rightarrow\infty}\sum_{k=0}^\infty a_k z_j^k=\sum_{k=0}^\infty a_k(-1)^k.$$
Thank you.
 A: I think that this is a classic result. First, replace $(-)^k a_k$ by $b_k$. We then have as hypothesis that the power series $\sum b_k$ is convergent. Our $z_j$ then has to converge to $1$, and if $D$ is the domain defined by $D=\{z;|z|\leq 1, |1-z|\leq 4(1-|z|)$, we want to show that if $f(z)=\sum b_k z^k$ and if $z_j\in D$, $z_j\to 1$ and $|z_j|<1$, then $f(z_j)\to f(1)$. This is done by showing that the convergence of the series $f$ is uniform on $D$, hence $f$ is continuous on $D$ (Note that $1\in D$). 
Let $\varepsilon>0$. There exists $N(\varepsilon)=N$ such that if $S_k$ is defined by $S_0=b_n$, $S_{k}=b_n+\cdots+b_{n+k}$, we have $|S_k|<\varepsilon$ for all $k$. Then for $m\geq n$
$$T_{m,n}(z)=\sum_{j=n}^{m}b_j z^j=S_0z^n+(S_1-S_0)z^{n+1}+\cdots+(S_{m-n}-S_{m-n-1})z^m$$
We rewrite this as
$$S_0(z^{n}-z^{n+1})+S_1(z^{n+1}-z^{n+2})+\cdots+S_{m-n-1}(z^{m-1}-z^{m})+S_{m-n}z^m$$
And then for $n\geq N$, $z\in D$, $|z|<1$:
$$|T_{m,n}(z)|\leq \varepsilon(|z|^{n}|1-z|+\cdots+|z|^{m-1}|1-z|)+\varepsilon |z|^m$$
We use that $\displaystyle 1+|z|+\cdots+|z|^{m-n-1}=\frac{1-|z|^{m-n}}{1-[z[}$, $[z|^{n}\leq 1$, $|z|^m\leq 1$ and $[1-z|\leq 4(1-|z|)$, and we see that $|T_{m,n}(z)|\leq 5\varepsilon$. For $z=1$, this inegality is obviously true, hence we have shown that the power serie is uniformly convergent in $D$, and we are done.
Added: Let $\displaystyle S_n(z)=\sum_{k=0}^n b_k z^k$. Then, for $m>n$, we have $S_m(z)-S_n(z)=T_{m, n-1}(z)$. Hence by the above, if $n\geq N+1$ we get $|S_m(z)-S_{n}(z)|=|T_{m,n-1}(z)|\leq 5\varepsilon$. Now let $m\to +\infty$. As $S_m(z)\to f(z)$, we get $\displaystyle |f(z)-S_{n}(z)|\leq 5\varepsilon$ for $n\geq N+1$. This show the uniform convergence on $D$ of the $S_n(z)$.  
