# Sum of reciprocals of perimeters of primitive Pythagorean triples

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)?

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.

Any help would be really appreciated. THANKS!

• For Problem 2, I'm fairly sure it exceeds 1/3, as I ran it through a few million primitive Pythagorean triples on Python, and it exceeded 1. Commented May 24, 2020 at 12:01
• @SharkyKesa Thanks for that, I personally used Excel to evaluate the sum of first 35 terms. However, this question appeared in a competition exam in Saudi Arabia, where only calculators such as fx-570es plus can be used. No Python nor Excel to be used. Commented May 24, 2020 at 12:05

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 $$\frac{\phi(n)}n=\prod_{p\mid n}\left(1-\frac1p\right)$$.

• Great, what about PROBLEM I? Can you help me plz? Commented May 24, 2020 at 13:45