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Thomas, Bruckner & Bruckner, Elementary Real Analysis.

Prove that for all r > 1,

$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$

So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$

$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$

Michael Hardy
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Malin
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3 Answers3

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Rinse and repeat the first line below $$\begin{align} {1\over \color{blue}{r}-1}&={1\over \color{blue}{r}+1}+{2\over \color{blue}{r}^2-1}\\ &={1\over r+1}+2\left({1\over\color{blue}{r^2}+1}+{2\over (\color{blue}{r^2})^2-1}\right)\\ &={1\over r+1}+\frac{2}{r^2+1}+\frac{4}{\color{blue}{r^4}-1}\\ &={1\over r+1}+\frac{2}{r^2+1}+4\left(\frac{1}{\color{blue}{r^4}+1}+\frac{2}{(\color{blue}{r^4})^2-1}\right)\\ &=\cdots \end{align}$$

Jack's wasted life
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For any $x\in(0,1)$ we have $$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)\cdots \tag{1} $$ that can be read as every positive integer has a unique representation in base-$2$.
By taking $\frac{d}{dx}\log(\cdot)$ of both sides, we get: $$ \frac{1}{1-x}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\cdots \tag{2} $$ and it is enough to evaluate the last identity at $x=\frac{1}{r}$ to prove the given claim, after trivial simplifications.

Michael Hardy
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Jack D'Aurizio
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Notice that

$$\frac 1 {r-1} - \left(\frac 1 {r+1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \left( \frac 1 {r-1} - \frac 1 {r+1} \right) - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \\ \frac 2 {r^2 - 1} - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \frac 4 {r^4 - 1} - \left(\frac 4 {r^4 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) \dots = \\ \frac {2^n} {r^{2^n} - 1} - \frac {2^n} {r^{2^n} + 1} = \frac {2^{n+1}} {r^{2^{n+1}} - 1} .$$

Since $r>1$ we now have that

$$\lim _{n \to \infty} \frac {2^{n+1}} {r^{2^{n+1}} - 1} = \lim _{t \to \infty} \frac t {r^t - 1} = 0 ,$$

which means that

$$\lim _{n \to \infty} \left( \frac 1 {r-1} - \left(\frac 1 {r+1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) \right) ,$$

i.e. that

$$\frac 1 {r-1} = \sum _{n=0} ^\infty \frac {2^n} {r^{2^n} + 1} .$$

Alex M.
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