TITLE: Series of the totient function
QUESTION [14 upvotes]: 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 ?

REPLY [8 votes]: 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 $$\sum_{n\leq x}\frac{(-1)^n}{\phi(n)}=\frac{105}{2\pi^{4}}\log x+O(1).$$