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Hello,

I am reading some material on circle method. Right now I am at its application to the binary Goldbach problem. To obtain a certain bound the fact $\sum_{n> X} 1/ \phi(n)^2 = O(1/X)$ is used. Could anyone please help me how to solve it?

Here $\phi$ is the Euler totient function. And I would like the result without $\epsilon$.

Thank you.

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    $\begingroup$ Why do you ask? $\endgroup$
    – Igor Rivin
    Jun 5, 2012 at 16:50
  • $\begingroup$ If I am wrong I apologize but this seems like a homework problem. $\endgroup$
    – Daniel Parry
    Jun 5, 2012 at 17:23

3 Answers 3

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Theorem 2.14 of "Multiplicative Number Theory" I. Classical Theory, by Montgomery & Vaughan implies that $$\sum_{n\leq x} \left(\frac{n}{\phi(n)}\right)^2 = O(x)$$

Use this and partial summation method with $$\sum_{n\leq x} \left(\frac{n}{\phi(n)}\right)^2 \frac{1}{n^2}$$

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Or, making different assumptions of what the question means: find a function $\phi$ so that $$ \sum_{n=X}^\infty \frac{1}{\phi(n)^2} = O(1/X)\qquad\text{as } X \to \infty $$ And in fact $\phi(n)=n$ is an example that satisfies that: $$ \sum_{n=X}^\infty \frac{1}{n^2} = \frac{1}{X}+\frac{1}{2X^2}+\frac{1}{6X^3}-\frac{1}{30X^5} + O(1/X^7) $$

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  • $\begingroup$ Whoever down-voted this: Gerald's answer is correct, and it is very hard to tell what the OP is actually asking, as Gerald points out. $\endgroup$
    – Igor Rivin
    Jun 5, 2012 at 20:02
  • $\begingroup$ My apologies for not being clear enough regarding my question. It is my first time on Mathoverflow and I am still trying to figure out how to use this properly... $\endgroup$
    – J Kasahara
    Jun 6, 2012 at 13:45
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From Theorem 7 in Pete Clark's notes it follows that you get $O(1/X^{1-\epsilon})$ for any $\epsilon > 0.$

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  • $\begingroup$ A downvote? Is the answer wrong? $\endgroup$
    – Igor Rivin
    Jun 5, 2012 at 20:02
  • $\begingroup$ Two downvotes. I am honored. $\endgroup$
    – Igor Rivin
    Jun 5, 2012 at 21:12

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