Does the sequence $S_n = \sum_{\substack{k \leq n \\ (k,n)=1}} \frac{1}{k}$ diverge as $n \to \infty$? We know that harmonic series
\begin{equation}
\sum_{n=1}^{\infty} \frac{1}{n}
\end{equation}
is a divergent series. Furthoremore, we have that
\begin{equation}
\sum_{p} \frac{1}{p},
\end{equation}
where $p$ is prime, diverges as well.
The claim I want to prove is that
\begin{equation}
\sum_{\substack{k \leq n \\ (k,n)=1}} \frac{1}{k}
\end{equation}
diverges, as $n \to \infty$. 
I have tried using comparison test, but unfortunately I could not obtain my claim.
 A: Think about what happens when $n$ is prime.
A: An exercise 2.1 #16 from Montgomery & Vaughan "Multiplicative Number Theory I" will be useful. The method presented here is what it requires to solve (b) of the exercise. 
Let $\mu(n)$ be the Mobius function, $\tau(n)$ be the number of divisors of $n$, and $w(n)$ be the number of distinct prime divisors of $n$. We apply the result that $\sum_{d|n}\mu(d)$ is $1$ if $n=1$,  and $0$ if $n>1$. 
We have for any $n\geq 1$, 
$$
\begin{align}
\sum_{k\leq n, (k,n)=1} \frac1k &= \sum_{k\leq n} \sum_{d|k, d|n}  \mu(d)\frac1k \ \ \ (\textrm{as} \ d|(k,n) \Longleftrightarrow d|k, d|n)\\
&=\sum_{d\leq n, d|n}\mu(d)\sum_{m\leq \frac nd} \frac1{dm}\ \ \ (\textrm{changing order of summation and put}k=dm )\\
&=\sum_{d|n}\frac{\mu(d)}d\sum_{m\leq \frac nd} \frac1m \ \ \ (\textrm{as} \ d\leq n \textrm{ can be dropped due to }d|n)\\
&=\sum_{d|n}\frac{\mu(d)}d \left(\log\frac nd + \gamma+O(\frac dn)\right) \ \ \ (\textrm{as} \ \sum_{m\leq x}\frac1m=\log x + \gamma + O(\frac1x))\\
&=\left(\log n + \gamma + \sum_{p|n}\frac{\log p}{p-1}\right) \frac{\phi(n)}n + O\left( \frac{2^{w(n)}}n\right) \ \ \ (\textrm{by 2.1.#16a})
\end{align}
$$
Therefore, for any $n\geq 1$, 
$$
\sum_{k\leq n, (k,n)=1} \frac1k \geq \frac{\phi(n)}n \log n + O\left(\frac{2^{w(n)}}n \right).
$$
There is an absolute constant $c>0$ such that $\phi(n)\geq n/c\log\log n$ for $n\geq 3$. Applying this, we obtain the lower bound
$$
\sum_{k\leq n, (k,n)=1} \frac1k \geq  \frac{\log n}{c\log\log n} + O\left(\frac{2^{w(n)}}n \right).
$$
By applying $2^{w(n)}\leq \tau(n) \leq 2\sqrt n$, there is an absolute constant $C>0$ such that for sufficiently large $n$, 
$$
\sum_{k\leq n, (k,n)=1} \frac1k \geq \frac{\log n}{c\log\log n} -\frac{2C}{\sqrt n}. 
$$
Thus, we obtain
$$
\lim_{n\rightarrow\infty}\sum_{k\leq n, (k,n)=1} \frac1k =\infty.
$$
