Divergence of reciprocal of primes, Euler

Question

On Wikipedia at link currently is:

\begin{align} \ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right) = -\sum_p \ln \left( 1-\frac{1}{p}\right) \\ & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\ & {} = \sum_{p}\frac{1}{p}+ \sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p ^3} + \sum_{p}\frac{1}{4p^4} + \cdots \\ & {} = \left( \sum_p \frac{1}{p} \right) + K \end{align}

And then Wikipedia says that $K<1$, without any explanation. How do we know that $$\sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p ^3} + \sum_{p}\frac{1}{4p^4} + \cdots$$ is equal to a constant $K<1$?

Answer 1

You can in fact do better and show that $K < 0.43$.

$$ \sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p ^3} + \sum_{p}\frac{1}{4p^4} + \cdots = \sum_{n=1}^{\infty}\sum_{p}\frac{1}{np^n} $$

$$ < \sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{nk^n} = \sum_{n=2}^{\infty}\frac{\zeta(n) - 1}{n} = 1 - \gamma \approx 0.422785 $$

where Gamma is the Euler-Mascheroni constant.

Answer 2

Hint :

$$\sum_p \frac {1}{p^s} < \sum_{n=1}^{\infty} \frac 1{n^s} = \zeta(s) $$

$\zeta(s)$ is Riemann Zeta function, and $\zeta(s)$ converges for each $s \in \Bbb R, s>1$