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