Sum of reciprocals of perimeters of primitive Pythagorean triples

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