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Is it possible to find an estimate of the summation $$s(n)=\sum_{k=1}^n\frac1{\varphi(k\cdot p_k)}$$ where $\varphi(n)$ is the totient function and $p_k$ the k-th prime?

The corresponding series seems to converge to the value $$\lim_{n\rightarrow\infty}s(n)=1.86491\ldots$$ but I don't see a simple way to prove it.

Many thanks.

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    $\begingroup$ $\frac{1}{\varphi(kp_k)}=\frac{1}{(p_k-1)\varphi(k)}$. Now use any reasonable lower bound on $\varphi(k)$, like this one $\endgroup$
    – Wojowu
    Commented May 26, 2021 at 20:59

1 Answer 1

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$\newcommand\vpi\varphi$It is known that $$\vpi(n)\ge\frac n{c\ln\ln(n+10)}=:\psi(n)$$ for some real $c>0$ and all natural $n$. Also, $\psi$ is increasing on the interval $[N,\infty)$ for some natural $N$. Therefore and because $p_k\ge k$ for all $k$, $$\sum_{k\ge N}\frac1{\vpi(kp_k)}\le\sum_{k\ge N}\frac1{\psi(kp_k)} \le\sum_{k\ge N}\frac1{\psi(k^2)}<\infty.$$

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