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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.

icecream_sandwich07
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1 Answers1

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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 $S_n(z)$.

Kelenner
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    You are welcome – Kelenner Jul 12 '16 at 08:29
  • I noticed that you did not use the conditions given in the question. Just wondering if it is because they are not useful. Also, how did you manage to prove f converges uniformly just by proving that $T_{m,n}(z)\leq5\epsilon$? – icecream_sandwich07 Jul 12 '16 at 12:15
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    The conditions are necessary, simply I have stated them a bit differently to simplify. I edit my answer to add some explanationss for the uniform convergence. just wait some minutes. – Kelenner Jul 12 '16 at 12:22
  • Thanks for your edit. I am much clearer now about the uniform convergence. Just to clarify in your edit, shouldn't $S_m(z)-S_n(z)=T_{m,n}(z)$ instead? Also, I do not understand how you got $|1-z|\leq4(1-|z|)$ from the given condition of $|1+z|\leq4(1-|z|)$. Other than that, you have been really helpful and i appreciate your time a lot. – icecream_sandwich07 Jul 14 '16 at 09:35
  • With the $b_k=(-1)^k a_k$, your series is first $\sum(-1)^k b_k z^k$. Now it is a series in $(-z)$. So we replace $z$ by $-z$, $z_j$ by $-z_j$, etc. The domain $D$ is then $|1-z|\leq 4(1-|z|)$. – Kelenner Jul 14 '16 at 09:41
  • Alright I got it! Thank you so much. :) – icecream_sandwich07 Jul 14 '16 at 11:49