Sum of reciprocals of perimeters of primitive Pythagorean triples

Question

The primitive Pythagorean triples, in increasing order of perimeters, are: $(3,4,5),(5,12,13),(8,15,17),\dots$

So, the perimeters of these triangles, in increasing order, are: $12,30,40,\dots$

Let $S_n$ be the sum of the reciprocals of the first $n$ perimeters. For example, $S_5=\frac{1}{12}+\frac{1}{30}+\frac{1}{40}+\frac{1}{56}+\frac{1}{70}=\frac{73}{420}$.

PROBLEM I: When expressing $S_2,S_3,S_4,\dots,S_{100}$ to their simplest fractions, for which value(s) of $n$ does $S_n$ have; the least numerator? the greatest numerator? the least denominator? the greatest denominator?

PROBLEM II: Does $S_\infty$ exist? If yes, what is its closed form (not necessary as a fraction)?

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For PROBLEM II: my trial (which is not a good way): I sum up the first $35$ terms, $S_\infty$ seems to approach $1/3$. I am not sure.

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Any help would be really appreciated. THANKS!

Answer 1

The sum of the reciprocals of the perimeters of the primitive Pythagorean triples diverges.

The primitive Pythagorean triples can be enumerated using Euclid’s formula; that is, there is a bijection between pairs of positive integers $n\lt m$ with $m+n$ odd and $\gcd(m,n)=1$ and primitive Pythagorean triples $\left(m^2-n^2,2mn,m^2+n^2\right)$. We have

\begin{eqnarray} \sum_{{\scriptstyle n\lt m\atop\scriptstyle \gcd(m,n)=1}\atop\scriptstyle 2\nmid m+n}\frac1{m^2-n^2+2mn+m^2+n^2} &=& \sum_{{\scriptstyle n\lt m\atop\scriptstyle \gcd(m,n)=1}\atop\scriptstyle 2\nmid m+n}\frac1{2m^2+2mn} \\ &\gt& \sum_{{\scriptstyle n\lt m\atop\scriptstyle \gcd(m,n)=1}\atop\scriptstyle 2\nmid m+n}\frac1{4m^2} \\ &\gt& \sum_{{\scriptstyle n\lt m\atop\scriptstyle \gcd(m,n)=1}\atop\scriptstyle 2\mid m}\frac1{4m^2} \\ &=& \frac14\sum_{2\mid m}\frac{\phi(m)}{m^2}\;, \end{eqnarray}

where $\phi(m)$ is Euler’s totient function. Since

$$ \liminf\frac{\phi(n)}n\log\log n=\mathrm e^{-\gamma} $$

(see Wikipedia), there is $n_0$ such that

$$ \frac{\phi(n)}n\gt\frac{\mathrm e^{-\gamma}}2\frac1{\log\log n} $$

for $n\ge n_0$. Then \begin{eqnarray} \sum_{\scriptstyle2\mid m\atop\scriptstyle m\ge n_0}\frac{\phi(m)}{m^2} &\gt& \frac{\mathrm e^{-\gamma}}2\sum_{\scriptstyle2\mid m\atop\scriptstyle m\ge n_0}\frac1{m\log\log m} \\ &\gt& \frac{\mathrm e^{-\gamma}}2\sum_{\scriptstyle2\mid m\atop\scriptstyle m\ge n_0}\frac1{m\log m}\;. \end{eqnarray}

Since $\int\frac{\mathrm dx}{x\log x}=\log\log x$, this sum diverges by the integral test.

Since we only used $\frac1{\log n}$ and not the tighter lower bound with $\frac1{\log\log n}$, you don’t need to know the exact limit inferior in the exam; a sufficiently good lower bound for $\frac{\phi(n)}n$ should be derivable from $\frac{\phi(n)}n=\prod_{p\mid n}\left(1-\frac1p\right)$.