Sum of reciprocal of primes If $p_n$ is the nth prime, then prove that:$$\frac1p_1+\frac1p_2+....+\frac1p_n$$ is not an integer.
Since $p_1 ,p_2,..., p_n$ are all prime their reciprocal is not an integer so the sum should not be an integer. But I am not sure.
 A: Hint: ${1\over p_1}+...+{1\over p_n}={{p_2..p_n+p_1p_3..p_n+..p_1..p_{n-1}}\over p_1..p_n}$.
$p_2..p_n+p_1p_3..p_n+..p_1..p_{n-1}$ is prime with $p_1$.
A: A more general result: 
if $a_1,\cdots,a_n$ are non-zero integers such that at least one of them does not divide the product of the others, then
$$\frac1a_1+\frac1a_2+\cdots+\frac1a_n$$
is not an integer.
WLOG we assume that $a_1$ does not divide $a_2\cdots a_n$.
Let $A_i=\prod_{j \neq i} a_j$, then
$$\frac1a_1+\frac1a_2+\cdots+\frac1a_n=\frac{A_1+\dots + A_n}{a_1\cdot a_2\cdots a_n}.$$
Now  $a_1$ divides the denominator and $A_2+\dots + A_n$, but $a_1$ does not divide $A_1$. Therefore $a_1$ does not divide the numerator and the fraction can not be an integer.
A: Hint: I would go for a proof by induction.  Note that for the sum to be an integer, we must have
$$
\frac 1{p_1} + \cdots + \frac 1{p_{n-1}} = k - \frac{1}{p_n}
$$
for some integer $k$.  Now, the number on the right can be written as a reduced fraction in the form $\frac{kp_n - 1}{p_n}$, so that its denominator is $p_n$.  Now, argue that the denominator of the rational number on the left (in reduced form) cannot also be $p_n$.
A: Suppose $$n=\frac1p_1+\frac1p_2+....+\frac1p_n \in \mathbb{Z}$$ 
Then we have $$n\prod_{i=1}^n p_i=\sum_i^n\prod_{j\neq i}p_j$$
Taking $\mod p_1$,
We have $$0 \equiv \prod_{j \neq 1} p_j \mod p_1$$
which is a contradiction.
A: Probably the simplest arithmetic gadget to describe the situation is that of a valuation. For a prime $p$, the $p$-adic valuation of a rational number $q$ is the exponent on $p$ in the prime factorization of $q$.
For example, $\frac{20}{21} = 2^2 3^{-1} 5^1 7^{-1}$, so $v_2(20/21) = 2$, $v_3(20/21) = -1$, $v_5(20/21) = 1$, $v_7(20/21) = - 1$, and $v_p(20/21) = 0$ for all other primes.
The key fact now is:

If $v_p(x) > v_p(y)$, then $v_p(x + y) = v_p(y)$

Using this, it is now very easy to compute that, for $m \leq n$
$$ v_{p_m} \left( \frac1p_1+\frac1p_2+....+\frac1p_n \right) = -1$$
and consequently, the sum can't be an integer.
A: To get the idea start with three distinct primes $p$, $q$, and $r$. If $\dfrac 1p  + \dfrac 1q + \dfrac 1r$ is an integer $k$ you have $$\dfrac 1p + \dfrac 1q + \dfrac 1r= k$$
so that $$pq + pr + qr = kpqr.$$
This implies, among other things, that $r | pq$ which isn't possible. This easily generalizes to $n$ distinct primes.
A: You can prove by induction that
$${1\over p_1}+{1\over p_2}+\cdots+{1\over p_n}={A_n\over p_1p_2\cdots p_n}\quad\text{with }p_k\not\mid A_n\text{ for }1\le k\le n$$
The base case, $p_1\not\mid1$, is obvious.  For the induction step,
$${1\over p_1}+{1\over p_2}+\cdots+{1\over p_n}+{1\over p_{n+1}}={A_n\over p_1p_2\cdots p_n}+{1\over p_{n+1}}={A_np_{n+1}+p_1p_2\cdots p_n\over p_1p_2\cdots p_np_{n+1}}$$
so it suffices to observe that $A_{n+1}=A_np_{n+1}+p_1p_2\cdots p_n$ is not divisible by any $p_k$ for $1\le k\le n+1$.  
Note, it is not necessary here for $p_k$ to be the $k$th prime; all that's needed is for the $p_k$'s to be distinct primes.
