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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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Why do you ask? – Igor Rivin Jun 5 '12 at 16:50
If I am wrong I apologize but this seems like a homework problem. – Daniel Parry Jun 5 '12 at 17:23
up vote 5 down vote accepted

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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Thank you very much! – J Kasahara Jun 6 '12 at 13:50

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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Whoever down voted this is an idiot. @Gerald's answer is correct, and it is very hard to tell what the OP is actually asking, as @Gerald points out. – Igor Rivin Jun 5 '12 at 20:02
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... – J Kasahara Jun 6 '12 at 13:45

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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A downvote? Is the answer wrong? – Igor Rivin Jun 5 '12 at 20:02
Two downvotes. I am honored. – Igor Rivin Jun 5 '12 at 21:12

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