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if $n>k>1$ be postive integer,show that

$$S_{k}(n)=\dfrac{1}{n^k}\sum_{j_{1}=1}^{n}\sum_{j_{2}=1}^{n}\cdots\sum_{j_{k}=1}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})\le\dfrac{\zeta(k-1)}{\zeta(k)} \tag{1}$$

where $\zeta(s)=\sum_{n=1}^{+\infty}\dfrac{1}{n^s},s>1$

I have known this $S_{2}(n)$some approximation reslut,such as following $$ \begin{align} S_{2}(n)&=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j) \\ &= \frac{1}{n^2}\sum_{g=1}^n\sum_{i\le\lfloor n/g\rfloor}\sum_{\substack{j\le\lfloor n/g\rfloor\\(i,j)=1}} g \\ &= \frac{1}{n^2}\sum_{g=1}^n g\left(-1+2\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i)\right) \\ &= -\frac{n(n+1)}{2n^2}+\frac{2}{n}\sum_{g=1}^n \frac{g}{n}\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i) \end{align}$$ Write $$ f(x) = \frac{1}{x}\sum_{i\le x}\varphi(i) = \frac{3x}{\pi^2}+E(x) \\ E(x) = o(\log x) $$ (see Eric Naslund's exposition) then $$ \begin{align} S_{2}(n) &= -\frac{1}{2}-\frac{1}{2n}+\frac{2}{n}\sum_{g=1}^{n}f(n/g) \\ &= -\frac{1}{2}-\frac{1}{2n}+\frac{6}{\pi^2}\sum_{g=1}^{n}\frac{1}{g}+\frac{2}{n}\sum_{g=1}^n E(n/g) \\ &= \frac{6}{\pi^2}\log n+\frac{6\gamma}{\pi^2}-\frac{1}{2}+C+o(1) \\ &= \frac{6}{\pi^2}\log n + C' + o(1) \end{align} $$ where the constant $C$ arises from $$ E(x) = o(\log x) \\ \left|\frac{2}{n}\sum_{g=1}^n E(n/g)\right|< \frac{C}{n}\sum_{g=1}^n\log(n/g)=C\left(\log n - \frac{\log n!}{n}\right)=C+o(1) $$ by Stirling's approximation. Calculations suggest $C=0.39344\cdots, C'=0.24434\cdots$. also see:2

But for $(1)$inequality, there exist some reslut?or anyone can help prove,Thanks

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    $\begingroup$ Just use $n=\sum_{d|n} \phi(d)$ and exchange sums. The inequality drops at once. $\endgroup$
    – Lucia
    Apr 22, 2020 at 3:45
  • $\begingroup$ Thanks, exchange the sums maybe is not easy to prove?Thanks $\endgroup$
    – math110
    Apr 22, 2020 at 3:48
  • $\begingroup$ No, it is easy to prove. Try, and throw out all that you wrote about the case k=2. And this doesn't seem like a research level problem -- where does it come from? $\endgroup$
    – Lucia
    Apr 22, 2020 at 3:55
  • $\begingroup$ oh,I don't understand,can you post your answer,Thanks,this problem is from web $\endgroup$
    – math110
    Apr 22, 2020 at 4:23

1 Answer 1

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Following Lucia's hint, we have that $$S_k(n)=\frac{1}{n^k}\sum_{d=1}^\infty\phi(d)\left\lfloor\frac{n}{d}\right\rfloor^k \leq\sum_{d=1}^\infty\frac{\phi(d)}{d^k} =\frac{\zeta(k-1)}{\zeta(k)}.$$

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  • $\begingroup$ Thanks,the last $=$ why? $\endgroup$
    – math110
    Apr 23, 2020 at 6:42
  • 1
    $\begingroup$ This is a standard fact about the Riemann zeta function. Look at the Euler product. $\endgroup$ Apr 23, 2020 at 10:53

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