I need a clarification on a simple proof

Question

Can somebody explain to me how in Sangchul Lee answer here: The set $\{\frac{\varphi(n)}n:n\in \Bbb N\}$ $|f(n_k) - f(n_{k-1})| < \epsilon$ implies for $f(n_k) \to 0$?

Answer 1

As I had previously commented (Saucy O'Path confirms) there is no such implication. On the other hand we can prove the result directly as follows:

I know it's a long shot, but, using the Euler Product formula $$\prod_{k=1}^\infty\left(1-\frac1{p_k^s}\right)=\frac1{\zeta(s)}$$ with $s=1$ and knowing that $\zeta(1)=\infty$ (this is the harmonic series) we obtain the result as,

$$\lim_{k\to\infty}f(n_k)=\prod_{k=\color{red}{N}}^\infty\left(1-\frac1{p_k}\right)$$ and the first terms (which we are missing) play no role in the product converging to $0$.