Series of the totient function

Question

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 ?

Answer 1

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 $$\sum_{n\leq x}\frac{(-1)^n}{\phi(n)}=\frac{105}{2\pi^{4}}\log x+O(1).$$

Answer 2

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.

Answer 3

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.