Swebaceacichegh

Answered

2022-01-21

Suppose $\left|x\right|<1$ . Can you give any ideas on how to find the following sum?

$\frac{x}{x+1}+\frac{2{x}^{2}}{{x}^{2}+1}+\frac{4{x}^{4}}{{x}^{4}+1}+\frac{8{x}^{8}}{{x}^{8}}+1+\dots$

Answer & Explanation

ma90t66690

Expert

2022-01-22Added 7 answers

Prove first that for any $x\in (-1,1)$ the identity

$\frac{1}{1-x}=\prod _{k\ge 0}(1+{x}^{{2}^{k}})$ (1)

follows from a telescoping product or the fact that every$n\in {\mathbb{N}}^{+}$ has a unique representation in base-2. By considering $\frac{d}{dx}\mathrm{log}(.)$ 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:

$\sum _{k\ge 0}\frac{{2}^{k}{x}^{{2}^{k}}}{1+{x}^{{2}^{k}}}=\frac{x}{1-x}$

follows from a telescoping product or the fact that every

from which:

Roman Stevens

Expert

2022-01-23Added 10 answers

Evaluating

$\sum _{q\ge 0}\frac{{2}^{q}{x}^{{2}^{q}}}{1+{x}^{{2}^{q}}}$

we obtain

$\sum _{q\ge 0}{2}^{q}\sum _{k\ge 0}{(-1)}^{k}{x}^{(k+1){2}^{q}}=\sum _{n\ge 1}{x}^{n}\sum _{{2}^{1}\mid n}{2}^{q}{(-1)}^{\frac{n}{{2}^{q}}-1}$

Now observe that

$\sum _{{2}^{q}\mid n}{2}^{q}{(-1)}^{\frac{n}{{2}^{q}}-1}=\sum _{p=0}^{{v}_{2}\left(n\right)}{2}^{p}{(-1)}^{\frac{n}{{2}^{q}}-1}$

where${v}_{2}\left(n\right)$ is the exponent of the highest power of 2 that divides n. This is

$\sum _{p=0}^{{v}_{2}\left(n\right)-1}{2}^{p}+{v}^{{2}^{{v}_{2}\left(n\right)}}-1)+{2}_{{v}_{2}\left(n\right)}=1$

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

$\sum _{n\ge 1}{x}^{n}=\frac{x}{1-x}$

we obtain

Now observe that

where

because

(This also goes through when n is is odd and we have one value for p, namelly zero.)

Hence the end result is

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