Proof that the sum of the reciprocals of primes diverges Reading the proof-sketch (as set out here) that the sum of the reciprocals of the prime numbers is divergent, I get stuck. The proof starts with Euler's product formula $\sum _{n=1}^{\infty } \frac{1}{n}=\prod_p \frac{1}{1-p^{-1}}$. Then it proceeds as follows:
"Euler considered the above product formula and proceeded to make a sequence of audacious leaps of logic. First, he took the natural logarithm of each side, then he used the Taylor series expansion for $\ln x$ as well as the sum of a converging series:
$$\ln \left(\sum _{n=1}^{\infty } \frac{1}{n}\right)=\ln \left(\prod_p \frac{1}{1-p^{-1}} \right)$$
$$=\sum_p \frac{1}{p} +\frac{1}{2p^2}+\frac{1}{3p^3}+...$$
$$=\sum_p \frac{1}{p}+\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+...$$
$$=A+\frac{1}{2}B+\frac{1}{3}C+...$$
$$=A+K$$
for a fixed constant $K<1$."
I don't follow the last step. Clearly, $B=\sum_p \frac{1}{p^2}$, $C=\sum_p \frac{1}{p^3}$, $D=\sum_p \frac{1}{p^4}$ and so on, and each of these sums is convergent. But why is $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D+...<1$ given that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ is divergent?
I get that the value of $\sum_p \frac{1}{p^k}$ becomes rapidly smaller with increasing $k$, but how does one prove the convergence of $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D+...$ to a value less than $1$?

Could be following from this (for $k\ge 2$): $$\sum_p\frac{1}{p^k}\lt\sum_{n=2}^{\infty}\frac{1}{n^k}\lt\int_1^{\infty}\frac{dx}{x^k}=\frac{1}{k-1}$$ so $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D\cdots$ becomes bounded by a simple telescopic series $\sum_{k=2}^\infty\frac{1}{(k-1)k}$ with the sum $1$?
