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I'd like to estimate the following sum $$ \sum_{n\leq x}\frac1{k_n}\;,\qquad x\rightarrow \infty\;, $$ where $k_n=[\mathbb{Q}(\zeta_n,a^{1/n}):\mathbb{Q}]$ is the degree of a Kummer extension for a fixed integer $a$, $a\neq 0,\pm1$. From the literature (Hooley's paper on Artin conjecture under GRH), we know that if $a=b^h$ for some integer $b=b_0 b_1^2$, with $b_0$ squarefree, being $h$ the maximum possible exponent, then $$ k_n=\frac{n\varphi(n)}{\delta(n)\gcd(n,h)}\;, $$ where $\delta(n)=1,2$ depending on $a$ (irrelevant for our asymptotic estimate).

Is it correct what follows? $$ \sum_{n\leq x}\frac1{k_n} \leq 2h \sum_{n\leq x}\frac1{n\varphi(n)} \ll \frac{\log x}{x}\;, $$ by Abel's summation formula.

Last but not least, does a similar result apply in the case $a\in\mathbb{Q}$? In this case, can we continue saying that $$ \sum_{n\leq x}\frac1{k_n}\leq C \sum_{n\leq x}\frac1{n\varphi(n)} $$ for some fixed constant $C$? That is, does the Hooley's computations of $k_n$ remain similar in this case? What's the analogy of $a=b^h=(b_0b_1^2)^h$ now? Any literature suggestion where I can find this argument explained?

Thanks in advance.

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  • $\begingroup$ how does $\delta$ depend on $a$ - just for fun - plus it might be relevant $\endgroup$ – JMP Oct 6 '15 at 20:21
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I can explain the first part. The second part I'm not sure about.

We have:

$$k_n=\dfrac{n\varphi(n)}{\delta(n)\gcd(n,h)}$$

Because $\delta(n)\le2$ so:

$$\sum_{n\leq x}\dfrac1{k_n} = \sum_{n\leq x}\dfrac{\delta(n)\gcd(n,h)}{n\varphi(n)}\le2\sum_{n\leq x}\dfrac{\gcd(n,h)}{n\varphi(n)}$$

Also $\gcd(n,h)<=h$, so:

$$\sum_{n\leq x}\dfrac1{k_n} \le2h\sum_{n\leq x}\dfrac1{n\varphi(n)}$$

Using Abel's Summation Formula with $a_n=\dfrac1n$ and $\phi(n)=\dfrac1{\varphi(n)}$

means that $A(x)\lt\log(x)$ and we also have that $\dfrac1{\varphi(x)}<\dfrac1x$, and so with the integral missing, we get:

$$\sum_{n\leq x}\dfrac1{k_n}\ll \frac{\log x}{x}$$

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    $\begingroup$ I'm wondering how a positive series could have a sum that tends to 0 asymptotically, though. $\endgroup$ – Fan Zheng Oct 7 '15 at 0:03
  • $\begingroup$ Hey so I think your inequality involving $\phi(x)$ and $x$ is backwards (I think you want to use $\phi(y) \gg y/\log\log{y}$), and also you are proving a bound on the sum over $n > x$ (else how can the sum go to zero? I may be confused by the definition but the other sum is at least $1/k_1$, no?). This way we get a bound of $\ll \sum_{n > x} \log\log{n}/n^2 \ll \log\log{x}/x$. $\endgroup$ – alpoge Oct 7 '15 at 1:01
  • $\begingroup$ i'm thinking about it - i like the sum $n>x$ bit $\endgroup$ – JMP Oct 7 '15 at 1:07
  • $\begingroup$ @JonMarkPerry: thanks, that's exactly what I did in the case of integer $a$. Maybe for the case $a$ rational I should post another specific question. Thanks again! $\endgroup$ – PITTALUGA Oct 7 '15 at 7:41

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