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.

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 $$\frac{z}{z-1}+\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}\frac{2^{n+1}z^{2^{n+1}}}{z^{2^{n+1}}-1}=0$$
