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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?

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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 $$p_1 p_2 \cdots p_n .$$

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+1 nice answer! –  draks ... Mar 22 '12 at 22:41
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