<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] Your approach is correct! It can be verified in the following way too.
Consider the partial sums $$S_n=\sum_{j=0}^{n}\frac{2^jz^{2^j}}{1+z^{2^j}}$$
Then $$\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}S_n$$
Observe (prove by induction on $n$) that $$S_n+\frac{z}{z-1}=\frac{2^{n+1}z^{2^{n+1}}}{z^{2^{n+1}}-1}$$
Now $$\frac{z}{z-1}+\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}\frac{2^{n+1}z^{2^{n+1}}}{z^{2^{n+1}}-1}=0$$
</s>