<s>[INST] Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$ Show that if $|z| < 1$, then
\begin{align*}
\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots
\end{align*}
converges
MY ATTEMPT
Let us start by noticing that
\begin{align*}
\frac{z}{1-z} - \frac{z}{1-z} + \frac{z}{1+z} = \frac{z}{1-z} + \frac{z - z^{2} -z - z^{2}}{1-z^{2}} = \frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}}
\end{align*}
Similarly, we have that
\begin{align*}
\frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}} + \frac{2z^{2}}{1+z^{2}} = \frac{z}{1-z} + \frac{2z^{2} - 2z^{4} - 2z^{2} - 2z^{4}}{1-z^{4}} = \frac{z}{1-z} - \frac{4z^{4}}{1-z^{4}}
\end{align*}
Based on this pattern, it can be proven by induction the given series is the same as
\begin{align*}
\frac{z}{1-z} = z + z^{2} + z^{3} + \ldots = \sum_{n=1}^{\infty}z^{n}
\end{align*}
which converges, since it is the geometric series inside the circle $|z| < 1$
(we may apply the ratio test, for instance).
Could someone tell me if my approach is correct? Any contribution is appreciated.
 [/INST] So far, you've found
$$
\sum_{k=0}^{n-1} \frac{2^{k}z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z} - \frac{2^{n}z^{2^{n}}}{1-z^{2^{n}}}.
$$
As you said, you'll need to prove this by induction. To finish the proof, you need to show
$$
\lim_{n\rightarrow\infty}\frac{2^n z^{2^n}}{1-z^{2^n}} =0.
$$
You can then combine these to get
$$
\lim_{n\rightarrow\infty}\sum_{k=0}^{n} \frac{2^{k}z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z},
$$
which proves convergence.
</s>