Need to prove formula for $\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$

Question

I need help showing that for $|z|<1$ we have

$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$ = $\frac{z}{1-z}$

I tried using $\sum_{n=0}^{\infty} (-z^{2^k})^{n} = \frac{1}{1+z^{2^k}}$ but that hasn't gotten me very far. Any help is appreciated!

Cauchy Condensation can help get bounds on your sum (and therefore show convergence), but won't get you equality yet. — Mark Schultz-Wu, Jun 13 '17 at 19:43
The problem before this had me show that for $|z|<1$ we have $\sum_{k=0}^{\infty} \frac{z^{2^{k-1}}}{1-z^{2^k}} = \frac{z}{1-z}$ I hope this also helps! — JSanchez, Jun 13 '17 at 19:50
Same thing I did above. I put $\frac{1}{1-z^{2^k}} = \sum_{n=0}^{\infty} (z^{2^k})^n$. From here you get a double sum of the form $\sum_{k=0}^{\infty}\sum_{n=0}^{\infty} z^{2^{n}(2k+1)}$ which equals $\sum_{k=1}^{\infty} z^k$. This doesn't help too much though because of the $2^k$ that appears in the new sum. — JSanchez, Jun 13 '17 at 19:56
Possible duplicate of Two different expansions of $\frac{z}{1-z}$ — rtybase, Jun 13 '17 at 21:40

score 3 · Answer 1 · answered Jun 13 '17 at 20:08

Let $$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}} = \sum_{k=0}^{\infty} a_n z^n$$ and examine contributions to the $a_n$ from the various terms in the $k$ sum.

For $n=0$ none of the terms in the sum contributes (the $k=0$ contribution is $z-z^2+z^3-z^4+\ldots$).

For $n$ odd only the $k=0$ term contributes, and thus for $n$ odd, $a_n = 1$.

The $k=1$ term of the sum contributes to $(-1)^{n/2+1}2^1$ to each $a_n$ for $n$ even. Higher $k$ terms give no contribution to $a_2$, or indeed to any $a_{4p+2}$, thus for $p\in\Bbb Z$ $$ a_{4p+2} = (-1) + 2 = 1 $$ The contributions of the $k=1$ term of the sum to each $a_{4p}$ is $-2$.

The $k=2$ term of the sum contributes to $(-1)^{n/4+1}2^2$ to each $a_n$ for $n$ divisible by $4$. Higher $k$ terms give no contribution to any $a_{8p+4}$, thus for $p\in\Bbb Z$ $$ a_{8p+4} = (-1) + (-2) + 4 = 1 $$ The contributions of the $k=2$ term of the sum to each $a_{8p}$ is $-4$.

The $k=3$ term of the sum contributes to $(-1)^{n/8+1}2^3$ to each $a_n$ for $n$ divisible by $8$. Higher $k$ terms give no contribution to any $a_{16p+8}$, thus for $p\in\Bbb Z$ $$ a_{16p+8} = (-1) + (-2) + (-4) + 8 = 1 $$ The contributions of the $k=2$ term of the sum to each $a_{8p}$ is $-8$.

Without going to the effort to formally introduce induction here, we can see the pattern: For any even $n>0$, the net contributions always add to $1$, and for odd $n$ we already saw a value of $a_n = 1$. Thus the result is
$$ \sum_{k=1}^\infty z^n = \frac{z}{1-z} $$

Thanks @MarkFischler nice idea! I will be adding that to my bag of tricks! — JSanchez, Jun 13 '17 at 20:40

Lutz Lehmann · Accepted Answer · 2017-06-14T05:37:51.343

Define $$ f(z)=\frac{z}{z-1}+\sum_{k=0}^\infty\frac{2^kz^{2^k}}{1+z^{2^k}} $$ then it is easy to confirm that $$ 2f(z^2)=f(z)\implies f(z)=2^kf(z^{2^k}) $$ Now it is a bit more tedious but still elementary to confirm that for all $r<1$ there is a constant $C_r$ so that for all $|z|<r$ $$ |f(z)|\le C_r |z|. $$ But then also $|z^{2^k}|<r$ and $2^k|z|^{2^k}\to 0$ for $k\to\infty$ which implies $$ f(z)=0 $$ for all $|z|<1$.

Another approach can use term-wise differentiation $$ \frac1{1-z}=\prod_{k=0}^\infty(1+z^{2^k})\\ -\ln(1-z)=\sum_{k=0}^\infty\ln(1+z^{2^k})\\ \frac1{1-z}=\sum_{k=0}^\infty\frac{2^kz^{2^k-1}}{1+z^{2^k}}\\ \frac{z}{1-z}=\sum_{k=0}^\infty\frac{2^kz^{2^k}}{1+z^{2^k}}\\ $$

(+1) for the second approach, which is clear and concise. – Mark Viola Jun 14 '17 at 17:09 — Mark Viola, Jun 14 '17 at 17:09

Need to prove formula for $\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$

2 Answers2