# Thread: Sum of inverses of primes

1. ## Sum of inverses of primes

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

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

2. Let $S=1/p_1+1/p_2+\cdots+1/p_n$. This leads to $p_1p_2p_3\cdots p_nS=p_2p_3\cdots p_n+p_1p_3\cdots p_n+\cdots+p_1p_2\cdots p_{n-1}$.

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

By the way, using this idea can show that $1/a_1+1/a_2+\cdots+1/a_n$ is not an integer if $(a_i,a_j)=1$ for $i\neq j$, i.e., the $a_i$ are relatively prime in pairs.

3. ## A different way, simpler maybe.

I didn't want to resist the following solution. Again let $S=1/p_1+1/p_2+\cdots+1/p_n$; suppose $S$ is an integer.
Each term in expasion of $p_2p_3\cdots p_nS$ is an integer except for $\displaystyle{\frac{p_2p_3\cdots p_n}{p_1}}$, but then $p_2p_3\cdots p_nS$ cannot be an integer.

4. Originally Posted by melese
I didn't want to resist the following solution. Again let $S=1/p_1+1/p_2+\cdots+1/p_n$; suppose $S$ is an integer.
Each term in expasion of $p_2p_3\cdots p_nS$ is an integer except for $\displaystyle{\frac{p_2p_3\cdots p_n}{p_1}}$, but then $p_2p_3\cdots p_nS$ 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