# "assuming that Pn is the nth prime number. establish that the sum 1/P1+1/P2+.....1/Pn is never an integer"

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f `p_n` is the `n^(th)` prime, prove that the sum `1/p_1+1/p_2+...+1/p_n` is never

an integer.

We proceed by induction:

(1) 1/2 is not an integer

(2) Assume for some `k>=2` that `sum_(i=1)^k1/(p_i)` is not an integer.

(3) We will show that `sum_(i=1)^(k+1)1/(p_i)` is not an integer. Now `sum_(i=1)^(k+1)1/(p_i)=sum_(i=1)^k1/(p_i)+1/(p_(k+1))` where `sum_(i=1)^k1/(p_i)` is not an integer. After possibly reducing this fraction we find that the numerator and denominator of this fraction have no common factors, and the denominator is `p_1*p_2*...*p_k`. Let this fraction be `a/(p_1*...*p_k)` where GCD`(a,p_1*...*p_k)=1` . Now we add `1/(p_(k+1))` .

The sum is `(p_(k+1)*a+p_1*p_2*...*p_k)/(p_1*p_2*...*p_(k+1))` .

Lemma: if a|(b+c) and a|b then a|c. Proof: if a|(b+c) then b+c=ma. If a|b then b=na. Then na+c=ma or c=a(m-n) so a|c.

Now if `(p_(k+1)*a+p_1*p_2*...*p_k)/(p_1*p_2*...*p_(k+1)) `

Consider the sum `p_(k+1)a+p_1*p_2*...*p_k` . Now `p_1*...*p_k` obviously divides `p_1*...*p_k` , but it cannot divide `(p_(k+1)) *(a)` .(By assumption above the fraction was reduced, and `p_(k+1)` is prime.) Then for the fraction to be an integer requires `p_(k+1)` to divide both terms in the numerator, since by the Lemma if it divides the sum it divides both terms. But `p_(k+1)` cannot divide `p_1*p_2*...*p_k` .

Therefore this fraction cannot be an integer.

Thus ` `the sum cannot be an integer by induction.