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Good morning,

I wonder if : $$\sum_{n} \frac{(-1)^n}{\varphi (n)}$$ converges or not.

where $\varphi (n)$ is the Euler function.

Do you have any idea ?

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    $\begingroup$ Well I think you can use a lower bound for $\varphi (n)$, I've seen a problem just like that just a few months ago, I will find it and tell you. $\endgroup$ – M.LTA Oct 13 '15 at 16:54
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    $\begingroup$ If the totient function were increasing, that series would be converging by Dirichlet's criterion. On the other hand, $\varphi(n)\geq\sqrt{n}$ for every $n\geq 7$ and $\varphi(n)$ essentially behaves like $\frac{n}{\log\log n}$, hence I think we may prove convergence through summation by parts, anyway. $\endgroup$ – Jack D'Aurizio Oct 13 '15 at 17:06
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    $\begingroup$ Chances are it will diverge. After all, the values of $\varphi(n)$ for even arguments are systematically lower than for neighboring odd. $\endgroup$ – Ivan Neretin Oct 13 '15 at 18:07
  • $\begingroup$ See OEIS A$211177$ and OEIS A$211178$. $\endgroup$ – Lucian Oct 13 '15 at 18:21
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Here is another way of proof (but still the same idea):

Since $\frac{-1()^n}{\varphi (n)}$ is bounded it is sufficient to prove that : $$S_{4n}:= \sum_{k=1}^n \left( -\frac{1}{\varphi (4k-3)}+\frac{1}{\varphi (4k-2)}-\frac{1}{\varphi (4k-1)}+\frac{1}{\varphi (4k)}\right)$$ does not converge.

To do this, just remark that $\varphi (4k-2)= \varphi (2k-1)$ so that : $$S_{4n}=A_n-B_n=\sum_{k=1}^n \frac{1}{\varphi (4k)} - \sum_{k=n}^{2n-1} \frac{1}{\varphi (2k+1)}$$

Since $\varphi (2m) \leqslant m$ we have $A_n \geqslant \frac{\log n }{2}$

We can prove the :

Lemma 1. If $n$ is odd then $\displaystyle \varphi (n) \geqslant 2n \frac{\log 3}{\log n} + 2 \log 3$

Proof. It is sufficient to write $n=p_1^{r_1} \dots p_d^{r_d}$ and note that $p_i \geqslant i+2$. Combined with the fact that $n \geqslant p_1 \dots p_d \geqslant 3^d$ the conclusion follows.

We obtain : $$B_n \leqslant \frac{\log 3}{2} \sum_{k=n}^{2n-1} \frac{\log (2k+1)}{2k+1}$ + 1 + \sum_{k=n}^{2n-1} \frac{1}{2k} + 1 \leqslant B'_n$$ with : $$B'_n = - \frac{1}{\log 3 \log n} + o((\log n)^{-1})$$

So that we can conclude.

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  • $\begingroup$ Very clear and useful, I like the lemma ! $\endgroup$ – user279923 Oct 13 '15 at 20:34
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It diverges, and in this answer we'll calculate exact asymptotics. We have that $$\sum_{n\leq x}\frac{(-1)^{n}}{\phi(n)}=\frac{\zeta(2)\zeta(3)}{3\zeta(6)}\log x+O(1)=\frac{105}{2\pi^{4}}\log x+O(1).$$

Write the sum as $$\sum_{n\leq x}\frac{(-1)^{n}}{\phi(n)}=\sum_{n\leq x}\frac{1}{\phi(n)}-2\sum_{\begin{array}{c} n\leq x\\ n\ \text{odd} \end{array}}\frac{1}{\phi(n)}.$$ Our goal will be to compare the sizes of $\sum_{\begin{array}{c} n\leq x\\ n\ \text{odd} \end{array}}\frac{1}{\phi(n)}$ and $\sum_{n\leq x}\frac{1}{\phi(n)}$. Define $$f(n)=\begin{cases} \frac{n}{\phi(n)} & \gcd(n,2)=1\\ 0 & \gcd(n,2)=2 \end{cases}.$$ Then we need to calculate $$\sum_{n\leq x}\frac{f(n)}{n}=\int_{1}^{x}\frac{1}{t}d\left(\sum_{n\leq t}f(n)\right)=\frac{1}{x}\sum_{n\leq x}f(n)+\int_{1}^{x}\frac{\sum_{n\leq t}f(n)}{t^{2}}dt.$$ Since $$\mu*f(p^{k})=\begin{cases} -1 & p=2,\ k=1\\ \frac{1}{p-1} & p\neq2,\ k=1\\ 0 & k\geq2 \end{cases},$$ using the methodology of this answer, since the factor of $\frac{3}{2}$ in the Euler product is replaced with a factor of $\frac{1}{2}$ at the prime $2$ we have that $$\sum_{n\leq x}f(n)=\frac{105}{2\pi^{4}}x+O(\log x),$$ and so $$\sum_{\begin{array}{c} n\leq x\\ n\ \text{odd} \end{array}}\frac{1}{\phi(n)}=\frac{105}{2\pi^{4}}\log x+O(1).$$ On the other hand, by the results of that same answer, we have that $$\sum_{n\leq x}\frac{1}{\phi(n)}=\frac{315}{2\pi^{4}}\log x+O(1)$$ and so $$\sum_{n\leq x}\frac{(-1)^n}{\phi(n)}=\frac{105}{2\pi^{4}}\log x+O(1).$$

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The sum diverges. Here is my reasoning.

$s =\sum_{n} \frac{(-1)^n}{\varphi (n)} =\sum_{n} \frac{1}{\varphi (2n)} -\sum_{n} \frac{1}{\varphi (2n+1)} =s_e-s_o $.

If $n$ is odd, $\phi(2n) =\phi(n) $.

If $n$ is even, $n = 2^ab$, $\phi(2n) =\phi(2^{a+1})\phi(b) =(2^{a}-1)\phi(b) $.

Therefore

$\begin{array}\\ s_e &=\sum_{n} \frac{1}{\varphi (2n)}\\ &=\sum_{n} \frac{1}{\varphi (4n-2)} +\sum_{k=2}^{\infty}\sum_n \frac{1}{\varphi (2^k 2n-1)}\\ &=\sum_{n} \frac{1}{\varphi (2n-1)} +\sum_{k=2}^{\infty}\sum_n \frac{1}{(2^{k-1}-1)\varphi ( 2n-1)}\\ &=\sum_{n} \frac{1}{\varphi (2n-1)}\left(1+\sum_{k=2}^{\infty}\frac{1}{2^{k-1}-1}\right)\\ &=s_o\left(1+c\right)\\ \text{where}\\ c &= \sum_{k=2}^{\infty}\frac{1}{2^{k-1}-1}\\ &> 0\\ \end{array} $

Therefore $s =s_e-s_o =c s_o $.

Since $s_o$ diverges (as shown in zhoraster's deleted answer which just considers the primes), $s$ diverges.

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  • $\begingroup$ The idea of the proof is very similar to mine, but both $s_e$ and $s_o$ are diverging series, so we have to be careful in manipulating them, and $s_e=s_o(1+c)$ has little meaning. $\endgroup$ – Jack D'Aurizio Oct 13 '15 at 19:40
  • $\begingroup$ Take the upper bounds in the sum to be $N$ and let $N$ get large. I am confident in my answer, although that doesn't mean much. $\endgroup$ – marty cohen Oct 13 '15 at 19:41
  • $\begingroup$ Exactly, we have to work with partial sums. The issue to overcome is that when we consider "twice that number" nothing grants we are still in the original interval, unless "that number" is in the first half of the previous interval. $\endgroup$ – Jack D'Aurizio Oct 13 '15 at 19:44

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