Convergence on variants of the Harmonic series

Question

I am aware that the harmonic series diverges. I am also aware that the series $$\sum_{n=1}^\infty\frac{1}{p_n}$$ diverges. But why does the series $$\sum\frac{1}{p_{p_n}}$$ converge?

Answer 1

The short answer is "because it can be proven". I'll try to give some insight.

Assuming you denoted the sequence of prim numbers by $p_n$:

$$\sum_{k=1}^\infty\frac{1}{p_{p_n}}$$ converges because it gets "small fast enough". By the prime number theorem, $p_n$ is, very very roughly, the same as $n\log n$ and $\sum\frac{1}{n\log n}$ does not converge. So to get an idea how $\sum\frac{1}{p_{p_n}}$ behaves, we could, again very very roughly, put $n\log n$ in $n\log n$ getting $$(n\log n)\log(n\log n)=n\log^2 n + n(\log n)\log^2 n$$

Now $\sum\frac{1}{n(\log n)\log^2 n}$ does not converge, but $\sum\frac{1}{n\log^2 n}$ does, so by comparison, the series $$\sum_{k=3}^{\infty}\frac{1}{n\log^2n+n(\log n)\log^2 n}$$ also converges and so does $\sum\frac{1}{p_{p_n}}$.