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Problem: Show that $$\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots $$ for all $r > 1$, with a hint given that $\displaystyle\frac {1} {r-1} - \frac {1} {r+1} = \frac {2} {r^2-1}$.

Thoughts: I am having difficulty seeing a connection from the hint to the problem. Any insights appreciated.

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    @anonymaker000010001 but $(r^2+1)$ does not factor over the reals. – 2017-02-01

2 Answers 2

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First, replace $r$ by $r^{2^n}$ in your identity to get:

$$\frac{1}{r^{2^n}-1}-\frac{1}{r^{2^n}+1}=\frac{2}{r^{2^{n+1}}-1}$$

Then consider the partial sum: $s_n=\frac{1}{r+1}+\frac{2}{r^2+1}+\ldots+\frac{2^{n-1}}{r^{2^{n-1}}+1}$

You have:

$$\begin{align}\frac{1}{r-1}-s_1&=\frac{1}{r-1}-\frac{1}{r+1}=\frac{2}{r^2-1}\\\frac{1}{r-1}-s_2&=\frac{1}{r-1}-s_0-\frac{2}{r^2+1}\\ &=\frac{2}{r^2-1}-\frac{2}{r^2+1}=\frac{4}{r^4+1}\end{align}$$

And so on, in general (by induction):

$$\frac{1}{r-1}-s_n=\frac{2^n}{r^{2^n}+1}$$

Then take the limit:

$$\frac{1}{r-1}-s=\lim_{n\to\infty}\frac{2^n}{r^{2^n}+1}=\lim_{u\to\infty}\frac{u}{r^u+1}=0$$ (with substitution $u=2^n$, and using $r>1)$. So $s=\frac{1}{r-1}$

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    "On both sides" refers to the hint's equation you mean? – 2017-02-01
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    No, to the identity you have to prove. – 2017-02-01
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    But doesn't that assume that they are equal in the first place? – 2017-02-01
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    I rewrote my answer in more detail. Hope it is clear now. – 2017-02-01
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    Ok I understand much better now thanks a lot! – 2017-02-01
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    i believe your first equation should have a minus not a plus. – 2017-02-01
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    Yes, i corrected it. – 2017-02-01
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    One more thing, what is $s_0$ – 2017-02-01
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    I only considered partial sums with at least one element. But you may take $s_0=0$ – 2017-02-01
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Note that if $r\neq1$, $\frac{n}{r^n-1}=\frac{n}{r^n+1}+\frac{2n}{r^{2n}-1}$ for any positive integer $n$.

Now, we can see taht $\frac{1}{r-1}=\frac{1}{r+1}+\frac{2}{r^2-1}=\frac{1}{r+1}+\frac{2}{r^2+1}+\frac{4}{r^4-1}=\cdots$

i.e. $\frac{1}{r-1}=\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}$.

Finally, for $r>1$, $\frac{1}{r-1}=\lim_{n\rightarrow\infty}\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}=\sum_{k=0}^{\infty}\frac{2^k}{r^{2^k}+1}$. Good luck.

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    This is really nice too actually well done. – 2017-02-01