Reciprocity of different prime numbers can approximate $1$?

Question

I want to see if there exist $p_1<p_2<p_3<\cdots<p_{1000}$ different prime numbers such that $|1/p_1+\cdots+1/p_{1000}-1|\le ({1\over p_{1000}})^2.$

a) what is my point with this? Nothing. But what is the point of the twin number conjecture? Maybe I can prove that such problem has always a solution, even for more terms.

b) what is my progress? We cannot have $|1/p_1+\cdots+1/p_{1000}-1|=0$ (which I proved), so there must be an error. Also, $|1/2+1/5+1/7+1/11+1/13-1|=0.0107<1/13$, so for five terms there is a solution with power 1, but 1/13^2 cannot be used here. In fact, I proved that the problem has a solution, if the expression on the right is simply $({1\over p_{1000}}),$ that is, the second power is just first power. (We can have $k$ terms, in general, too, not just 1000 terms.)

I wish to have second power (or anything bigger than 1 is great as long as we tend to infinity with the number of terms), because of Hurwitz theorem

https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory)

Remark: the sum of repciprocal of all primes is infinity, that looks good/promising in order not to get a contradiction.

Answer 1

Take the primes from $23$ to $7993$, remove prime $337$ and append the primes $8101$ and $33409$. In PARI/GP, you can create this vector as follows :

? s=primes([23,7993]);s=setminus(s,[337]);s=concat([s,8101,33409]);print(Set(isp
rime(s,2))==[1],"   ",vecsort(s)==s,"   ",Set(s)==s,"   ",length(s),"     ",abs(
sum(j=1,length(s),1/s[j])-1)<1/s[1000]^2)
1   1   1   1000     1
?

The output also shows that $s$ does the job : All entries are (proven) primes, the entries are ordered strictly increasing and there is no duplicate, the length is $1000$ and the sum also satisfies the given inequality.