Resolved: "Show that∑n=1∞z2n1−z2n+1is algebraic."

Zaynah Dunn

Answered question

2022-02-24

Show that

\sum_{n = 1}^{\infty} \frac{z^{2^{n}}}{1 - z^{2^{n + 1}}}

is algebraic.

Zernerqcw

Beginner2022-02-25Added 11 answers

There is in fact a nice closed form for the partial-sums that you can find just by calculating a bit:

f (z) = \frac{z^{2}}{1 - z^{4}} + \frac{z^{4}}{1 - z^{8}} + \frac{z^{8}}{1 - z^{16}} + \dots

= \frac{z^{2} (1 + z^{4}) + z^{4}}{1 - z^{8}} + \frac{z^{8}}{1 - z^{16}} + \dots

= \frac{z^{2} + z^{4} + z^{6}}{1 - z^{8}} + \frac{z^{8}}{1 - z^{16}} + \dots

= \frac{(z^{2} + z^{4} + z^{6}) (1 + z^{8}) + z^{8}}{1 - z^{16}} + \dots

= \frac{z^{2} + z^{4} + z^{6} + z^{8} + z^{10} + z^{12} + z^{14}}{1 - z^{16}} + \dots

So first few partial sums are given by

f_{1} (z) = \frac{z^{2}}{1 - z^{4}} = \frac{z^{2} - z^{4}}{1 - z^{2}} \frac{1}{1 - z^{4}}

f_{2} (z) = \frac{z^{2} + z^{4} + z^{6}}{1 - z^{8}} = \frac{z^{2} - z^{8}}{1 - z^{2}} \cdot \frac{1}{1 - z^{8}}

f_{3} (z) = \dots = \frac{z^{2} - z^{16}}{1 - z^{2}} \cdot \frac{1}{1 - z^{16}}

It should then be easy to show by induction that

f_{n} (z) = \frac{z^{2}}{1 - z^{2}} \cdot \frac{1 - z^{2^{n - 1}}}{1 - z^{2^{n}}}

The second factor is

1 + O (z^{2^{n - 1}}) \to 1

, as

n \to \infty

, thus

f

Do you have a similar question?Recalculate according to your conditions!

Didn't find what you were looking for?