Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$

Question

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.

Answer 1

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

Answer 2

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.