A finite sum of prime reciprocals How can you prove that $\sum\limits_k \frac1{p_k}$, where $p_k$ is the $k$-th prime, does not result in an integer?
 A: Inspiring by this: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes#Partial_sums
$\displaystyle \frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_n} = \frac{1}{2} + \frac{1}{odd} + \cdots + \frac{1}{odd} = \frac{odd.odd \dots odd + 2.odd \dots odd + \cdots + 2.odd \dots odd}{2.odd \dots odd}$
$\displaystyle = \frac{odd + even + \dots + even}{2.odd} = \frac{odd + even}{even} = \frac{odd}{even} \neq integer$.
A: I am assuming that the question means finite sums. (For any desired $s \in \mathbb R_{> 0}$, there exist infinite subsequences of the primes such that the corresponding series sums to $s$.)
Assume that for some primes $p_1, p_2, \ldots, p_n$, the quantity
$$
s = \frac{1}{p_1} + \frac{1}{p_2} + \cdots+ \frac{1}{p_n} 
$$
is an integer. Multiplying by $Q = p_2 p_3 \cdots p_n$ and rearranging, we note that
$$
\frac{Q}{p_1} = sQ - \frac{Q}{p_2} - \frac{Q}{p_3} - \cdots - \frac{Q}{p_n}.
$$
Each term on the right hand side is an integer (justify this!), and so the right (and hence the left) hand side itself is an integer. That is, $p_1$ divides $Q = p_2 p_3 \cdots p_n$. Do you see how this is a contradiction? 
Now, by a simple modification, one can show that (Exercise!) the denominator of the sum is exactly $$p_1 p_2 \cdots p_n .$$
