Sum of reciprocal of primes

Question

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.

Answer 1

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 $p_1$.

Answer 2

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 $p_n$.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.