Swebaceacichegh

2022-01-21

Suppose $|x|<1$. Can you give any ideas on how to find the following sum?
$\frac{x}{x+1}+\frac{2{x}^{2}}{{x}^{2}+1}+\frac{4{x}^{4}}{{x}^{4}+1}+\frac{8{x}^{8}}{{x}^{8}}+1+\dots$

ma90t66690

Expert

Prove first that for any $x\in \left(-1,1\right)$ the identity
$\frac{1}{1-x}=\prod _{k\ge 0}\left(1+{x}^{{2}^{k}}\right)$ (1)
follows from a telescoping product or the fact that every $n\in {\mathbb{N}}^{+}$ has a unique representation in base-2. By considering $\frac{d}{dx}\mathrm{log}\left(.\right)$ of both sides of (1), we get
$\frac{1}{1-x}=\sum _{k\ge 0}\frac{{2}^{k}{x}^{{2}^{k}-1}}{1+{x}^{{2}^{k}}}$
from which:
$\sum _{k\ge 0}\frac{{2}^{k}{x}^{{2}^{k}}}{1+{x}^{{2}^{k}}}=\frac{x}{1-x}$

Roman Stevens

Expert

Evaluating
$\sum _{q\ge 0}\frac{{2}^{q}{x}^{{2}^{q}}}{1+{x}^{{2}^{q}}}$
we obtain
$\sum _{q\ge 0}{2}^{q}\sum _{k\ge 0}{\left(-1\right)}^{k}{x}^{\left(k+1\right){2}^{q}}=\sum _{n\ge 1}{x}^{n}\sum _{{2}^{1}\mid n}{2}^{q}{\left(-1\right)}^{\frac{n}{{2}^{q}}-1}$
Now observe that
$\sum _{{2}^{q}\mid n}{2}^{q}{\left(-1\right)}^{\frac{n}{{2}^{q}}-1}=\sum _{p=0}^{{v}_{2}\left(n\right)}{2}^{p}{\left(-1\right)}^{\frac{n}{{2}^{q}}-1}$
where ${v}_{2}\left(n\right)$ is the exponent of the highest power of 2 that divides n. This is
$\sum _{p=0}^{{v}_{2}\left(n\right)-1}{2}^{p}+{v}^{{2}^{{v}_{2}\left(n\right)}}-1\right)+{2}_{{v}_{2}\left(n\right)}=1$
because $\frac{n}{{2}^{p}}$ is even unless $p={v}_{2}\left(n\right)$
(This also goes through when n is is odd and we have one value for p, namelly zero.)
Hence the end result is
$\sum _{n\ge 1}{x}^{n}=\frac{x}{1-x}$

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