# Sum of inverses of primes

• Sep 6th 2010, 03:14 PM
kimberu
Sum 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 PM
melese
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 AM
melese
A 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 AM
tonio
Quote:

Originally Posted by melese
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.

This is basically the very same nice solution you first posted, but with the conclusion shown in a slightly different way.

Tonio