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I am currently reading Vaughan's book 'The Hardy-Littlewood method' and am working on exercise 3.3 (page 37). I have tried working through a special case but that didn't help either. More concretely, I failed to show that

\begin{align} \sum_{q=1}^{\infty} \frac{(-1)^{q+1}\mu(q)}{\phi(q)^2} \ge 1, \end{align} , here $\phi$ is the Euler-Phi function and $\mu$ is the Möbius function. I have tried taking the first two terms (which form the bulk of the sum) and then estimate the rest of the sum using Cauchy-Schwarz and the lower bound for $\phi$, but it gets very messy. Is there a pretty solution? (Some numerical calculations suggest that the sum converges to $\sim 1.3$)

Edit: Proving convergence is easy, just use that $\phi(n) \gg n^{1-\delta}$ for all $\delta>0$. The difficulty is now in calculating the exact value (or approximate it in some way).

We note that we can also write our sum as (using absolute convergence and Euler factorization since $\mu(q)$ and $\phi(q)$ are multiplicative.

\begin{align} &\sum_{q \text{ odd }} \frac{\mu(q)}{\phi(q)^2}- \sum_{q \text{ even }} \frac{\mu(q)} {\phi(q)^2}, \\&=\prod_{p>2} \left(1+\frac{\mu(p)}{\phi(p)^2}+\frac{\mu(p^2)}{\phi(p^2)^2} +\cdots \right)-\frac{\mu(2)}{\phi(2)} \prod_{p>2} \left(1+\frac{\mu(p)}{\phi(p)^2}+\frac{\mu(p^2)}{\phi(p^2)^2} +\cdots \right),\\ &=\prod_{p>2} \left(1-\frac{1}{(p-1)^2} \right)+ \prod_{p>2} \left(1-\frac{1}{(p-1)^2} \right),\\ &=2 \prod_{p>2} \left(1-\frac{1}{(p-1)^2} \right), \end{align} This looks alot more managable already, now I would like to do something as in https://math.stackexchange.com/a/84721/45878.

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Pol van Hoften
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    Why "...that is, that it converges"? – Timbuc Dec 23 '14 at 13:08
  • Well, I expect it to converge. – Pol van Hoften Dec 23 '14 at 13:18
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    Yes, me too. Perhaps you can try some asymptotic estimation: $;\varphi(n)\sim n;$ for big $;n;$ , so $;\frac1{\varphi(n)^2}\sim\frac1{n^2};$ and you can then apply some variation of Dirichlet's Test. – Timbuc Dec 23 '14 at 13:34
  • I dont think that works, a lower bound of the form $\phi(n) \ge c n$ and so $\tfrac{1}{\phi(n)^2}\le \tfrac{c^2}{n^2}$ would be great but I dont think it holds. – Pol van Hoften Dec 23 '14 at 13:43
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    that might be impossible to achieve as $$\frac{\varphi(n)}{n^{1-\epsilon}}\xrightarrow[nt\to\infty]{}\infty;,;;; \forall;\epsilon>0$$ – Timbuc Dec 23 '14 at 13:46
  • I think proving convergence is easy (using that $\phi(n)\gg n^{2/3}$ for example). Actually calculating the constant (which is what I want if it wasn't clear) might be harder, but I think I have an Euler Product approach that works now. – Pol van Hoften Dec 23 '14 at 13:59

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Well spotted that splitting into odd and even terms simplifies things a lot.

To show that the sum of the series is $\geqslant 1$ we can use a fairly brutal estimate: $$\prod_{p > 2}\biggl(1 - \frac{1}{(p-1)^2}\biggr) > \prod_{k > 2}\biggl(1 - \frac{1}{(k-1)^2}\biggr) = \prod_{m = 2}^{\infty} \biggl(1 - \frac{1}{m^2}\biggr) = \frac{1}{2}\,.$$ A less brutal estimate using that all primes $> 2$ are odd gives $$\prod_{p > 2}\biggl(1 - \frac{1}{(p-1)^2}\biggr) > \prod_{m = 1}^{\infty} \frac{4m^2-1}{4m^2} = \frac{2}{\pi}$$ using the Wallis product.

The actual value of the product is known as the twin prime constant $$C_2 = \prod_{p > 2}\biggl(1 - \frac{1}{(p-1)^2}\biggr) \approx 0.66016181584686957392781211001455577843\,,$$ so the sum of the series is $2C_2 \approx 1.32032363169373914785562422002911155686$.

Apparently there is no simpler expression for $C_2$ known, in particular $C_2$ has no (simple) expression in terms of values of the $\zeta$-function.

Daniel Fischer
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