Series of the totient function 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 ?
 A: 
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 $$\sum_{n\leq x}\frac{(-1)^n}{\phi(n)}=\frac{105}{2\pi^{4}}\log x+O(1).$$
A: 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.
A: 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.
