Infinite sum of each term of the harmonic series times a corresponding order p-series Here is a proof from Wikipedia's article on the divergence of the infinite sum of the reciprocals of primes:
\begin{aligned}\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 + \frac 1 2 \sum_p \frac 1 {p^2} + \frac {1}{3} \sum_p \frac 1 {p^3} + \frac {1}{4} \sum_p \frac 1 {p^4} + \cdots \\
&{}=A+\frac {1}{2} B + \frac {1}{3} C + \frac {1}{4} D+\cdots \\[5pt]
&{}=A+K\end{aligned}
I follow this reasoning, however it then concludes that $K \le 1$. How is this determined. I have seen a general form of the nth order p-series, however it's derivation required real analysis or fourier transforms or something, stuff way over my head. Is there a simpler justification for this?
 A: Note that $$\sum_{n=2} \sum_{p}\frac{1}{np^n} = \sum_{p} \sum_{n=2} \frac{1}{np^n}\le \sum_{p}  \sum_{n=2}\frac{1}{2p^n} = \frac{1}{2}\sum_{p}\frac{\frac{1}{p^2}}{1-\frac{1}{p}}= \frac{1}{2}\sum_{p}\frac{1}{p(p-1)} \le \frac{1}{2}\sum_{n\ge2}\frac{1}{n(n-1)}  \le \frac{1}{2}\cdot1 <1$$
In the first step, I reversed the order of summation.  
In the second step, I used the fact that for $n\ge2$, $\frac{1}{n} \le \frac{1}{2}$.
In the third step, I used the fact that $\sum_{k=2}^\infty (1/p)^k = \frac{\frac{1}{p^2}}{1-\frac{1}{p}}$.
In the fourth step, I used the fact that the terms of the series are positive, and hence the summation over the primes is less than or equal to the summation over the natural numbers.  
In the final step, I used the fact that the series is telescoping.
A: Note that:
$$\begin{align}\sum_{k=2}^\infty\sum_p\frac1{kp^k}&=\sum_p\sum_{k=2}^\infty\frac1{kp^k}\\&=\sum_p\left(-\ln\left(1-\frac1p\right)-\frac1p\right)\\&<\sum_{n=2}^\infty\left(-\ln\left(1-\frac1n\right)-\frac1n+\frac4{5n^2}\right)\\&=\frac{2\pi^2}{15}-\frac15-\gamma\\&\approx0.9387\\&<1\end{align}$$
where I used
$$\ln(\Gamma(x))+(\gamma-1)x=\sum_{n=2}^\infty\frac xn-\ln\left(1+\frac xn\right)$$
