<s>[INST] Question about Euler proof of divergence of the sum of the reciprocals of the primes. Consider the Euler proof : 
$$\ln\sum_{n = 1}^{\infty}\frac{1}{n} = -\ln\sum_{p}\frac{1}{1-p^{-1}} = \sum_{p}\left(\frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \dots\right) = A + \frac{1}{2}B + \frac{1}{3}C + \dots = A + K,$$ where $K < 1$. 
My question is : why $K < 1$. Is it true because of $\frac{1}{2}B < \frac{1}{2}$, then $\frac{1}{3}C < \frac{1}{4}$  etc. So we could estimate our residue using sum of power of two?
 [/INST] Yes, you are right. For all $d\geq 2$
$$\frac{1}{d}\sum_p\frac{1}{p^d}<\frac{1}{2^{d-1}}.$$
Infact, for $d\geq 2$,
$$\sum_p\frac{1}{p^d}\leq \frac{1}{2^d}+\sum_{n\geq 3}\frac{1}{n^d}< \frac{1}{2^d}+\int_{x\geq 2}\frac{dx}{x^d}= \frac{1}{2^d}+\frac{2}{(d-1)2^d}\leq \frac{3}{2^{d}}$$
which implies that
$$\frac{1}{d}\sum_p\frac{1}{p^d}\leq\frac{3}{d2^d}<\frac{1}{2^{d-1}}.$$
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