How can we find an asymptotic formula for $\sum_{\substack{1\leq k\leq n \\ (n,k)=1}}f(k)?$ Here $f$ is some function and $(n,k)$ is the gcd of $k$ and $n$. I am particularily interested in the case $\sum_{\substack{1\leq k\leq n \\ (n,k)=1}}\frac{1}{k}.$ I know about the result $\sum_{\substack{1\le k\le n\\(n,k)=1}}k=\frac{n\varphi(n)}{2}$ which was discussed here, but I don't know if I can use it in the case of $f(k)=1/k$.
Asymptotics for sums of the form $\sum \limits_{\substack{1\leq k\leq n \\ (n,k)=1}}f(k)$
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0You can try Dirichlet generating function: let \tilde G(z)=\sum_{n>0}n^{-z}, we have \zeta(z+1)\tilde G(z)=\sum_{n>0}n^{-z}H_n. It might be useful in analytic number theory. – 2012-06-29
1 Answers
Hint: Try using the fact that $\sum_{d|n} \mu(d)$ is an indicator function for when $n=1$. This allows us to do the following for any function $f$:
$\sum_{n\leq x}\sum_{k\leq n,\ \gcd (k,n)=1} f(k,n)=\sum_{n\leq x}\sum_{k\leq n} f(k,n) \sum_{d|k, \ d|n} \mu (d) =\sum_{d\leq x} \mu(d) \sum_{n\leq \frac{x}{d}}\sum_{k\leq n} f(dk,nk).$
This method is very general, and works in a surprisingly large number of situations. I encourage you to try it.
Remark: Using this approach I get $\sum_{n\leq x}\sum_{k\leq n,\ \gcd(k,n)=1} \frac{1}{k}=\frac{6x}{\pi^{2}}\log x+\left(-\frac{\zeta^{'}(2)}{\zeta(2)^2}+\frac{6\left(\gamma-1\right)}{\pi^{2}}\right)x+O\left(\log^{2}x\right).$
Edit: I made a slight miscalculation in my remark, missing the factor of $\zeta(2)^2$ in the $\zeta^{'}(2)$ term, and have updated the asymptotic.
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0I think it should be $f(dk,dn)$ instead of $f(dk,nk)$ in the end? – 2012-10-30