Assuming $\displaystyle p_n$ is the nth prime number, establish that the sum $\displaystyle 1/p_1 + 1/p_2 + ... + 1/p_n $ is never an integer.

I don't know how to go about showing this. Any help is greatly appreciated, thanks!

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- Sep 6th 2010, 03:14 PMkimberuSum of inverses of primes
Assuming $\displaystyle p_n$ is the nth prime number, establish that the sum $\displaystyle 1/p_1 + 1/p_2 + ... + 1/p_n $ is never an integer.

I don't know how to go about showing this. Any help is greatly appreciated, thanks! - Sep 6th 2010, 05:17 PMmelese
Let $\displaystyle S=1/p_1+1/p_2+\cdots+1/p_n$. This leads to $\displaystyle p_1p_2p_3\cdots p_nS=p_2p_3\cdots p_n+p_1p_3\cdots p_n+\cdots+p_1p_2\cdots p_{n-1}$.

Assume to the contrary that $\displaystyle S $ is an integer; then $\displaystyle p_1 $ divides $\displaystyle p_2p_3\cdots p_n+p_1p_3\cdots p_n+\cdots+p_1p_2\cdots p_{n-1}$. Try to find a contradiction...

By the way, using this idea can show that $\displaystyle 1/a_1+1/a_2+\cdots+1/a_n $ is not an integer if $\displaystyle (a_i,a_j)=1 $ for $\displaystyle i\neq j$, i.e., the $\displaystyle a_i $ are relatively prime in pairs. - Sep 7th 2010, 12:16 AMmeleseA different way, simpler maybe.
I didn't want to resist the following solution. Again let $\displaystyle S=1/p_1+1/p_2+\cdots+1/p_n$; suppose $\displaystyle S $ is an integer.

Each term in expasion of $\displaystyle p_2p_3\cdots p_nS$ is an integer except for $\displaystyle \displaystyle{\frac{p_2p_3\cdots p_n}{p_1}} $, but then $\displaystyle p_2p_3\cdots p_nS$ cannot be an integer. - Sep 7th 2010, 02:33 AMtonio