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I am interested in the rate of decay of the sum

$$\sum_{n=N}^{\infty}\frac{\phi(n)}{n^{s}}$$ where $\phi$ is Euler's totient function and $s>2$ (in which case the sum converges trivially).
Of course there's the trivial bound $\sum_{n=N}^{\infty}\frac{\phi(n)}{n^{s}}\leq\sum_{n=N}^{\infty}\frac{1}{n^{s-1}}\leq C(s)\frac{1}{N^{s-2}}$.
I'm wondering if it's possible to do any better.

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    $\begingroup$ You can use the lower bound $\phi(n) \geq C \frac{n}{\log \log n}$ to get the other direction. Here you can take $C \approx e^{-\gamma}$ for large $n$. So in particular, you cannot get any better power saving. $\endgroup$
    – Lior Bary-Soroker
    Apr 17, 2018 at 14:45
  • $\begingroup$ Some improvement in the constant can be had by looking at n a multiple of small primes. Big primes don't help. Gerhard "Not Seeing Any Power Saving" Paseman, 2018.04.17. $\endgroup$
    – Gerhard Paseman
    Apr 17, 2018 at 15:58

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It is impossible to do any better. We have

$$\sum_{X\leq n<2X} \varphi(n)=\frac{9}{\pi^2}X^2+O(X \log X)$$

Therefore, for any positive $s>2$ we have

\begin{align*} \sum_{n\geq N} \frac{\varphi(n)}{n^s} &\geq \sum_{k\geq 0} 2^{-(k+1)s}N^{-s}\sum_{2^kN\leq n<2^{k+1}N} \varphi(n) \\ &\geq C\sum_{k\geq 0} 2^{-(k+1)s}N^{-s}2^{2k}N^2\\ &\geq C_0(s)N^{2-s} \end{align*} for some positive constants $C$ and $C_0(s)$.

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    $\begingroup$ Indeed, with a more careful version of this argument (partial summation) one can even get the asymptotic $$\sum_{n\geq N} \frac{\varphi(n)}{n^s} \sim \frac6{\pi^2} \frac{N^{2-s}}{s-2} \qquad(s>2).$$ $\endgroup$
    – Greg Martin
    Apr 17, 2018 at 17:12

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