# Is there are similar conjecture like this??

Talking with my friend, my friend suggest impressive conjecture that

For $i\in\mathbb{N}$, there are always exist natural number $r$ that satisfies $$\sum_{n=1}^{i} \frac{1}{n^r}=\frac{p}{q} , \gcd(p,q)=1$$ and $p+q$ is a prime number.

For example, $$\sum_{n=1}^{2} \frac{1}{n}=\frac{3}{2}$$ and $3+2=5$. And $i=3$, $$\sum_{n=1}^{3} \frac{1}{n^4}=\frac{1393}{1296}$$ and $1393+1296=2689$.

$$\sum_{n=1}^{4} \frac{1}{n^2}=\frac{205}{144}$$ and $205+144=349$ which is 70th prime number.

In $i=5$ $$\sum_{n=1}^{5} \frac{1}{n^3}=\frac{256103}{216000}$$ and $256103+216000=472103$.

Also $$\sum_{n=1}^{6} \frac{1}{n^3}=\frac{28567}{24000}$$ and $24000+28567=52567$.

There are similer like this? If not, What about your think, is it true? Or false....

Note: ID:metamorphy suggest that the sequence of the smallest values of $r$ begins with $1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 34, 1, 1, 5, \ldots$. (There are full comment in bellow with more comments.)

• I think you must mean "For every positive integer $i$". Sep 26 '18 at 14:59
• @TonyK yes,i'll fix it Sep 26 '18 at 14:59
• Funny. The sequence of the smallest values of $r$ begins with $1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 34, 1, 1, 5, \ldots$ (here $r_{13} = 34$). I couldn't find $r_{17}$... Sep 26 '18 at 15:29
• I don't know. If $r_{17}$ exists at all, then it is $> 1000$ (my computer was running a simple PARI program for $\approx 5$ minutes to get this). Sep 26 '18 at 15:44
• Feel free. I would also append the question on $r_{17}$. The PARI "program" is $\texttt{foo(n)={for(r=1,+oo,s=sum(k=1,n,k^(-r));if(isprime(numerator(s)+denominator(s)),return(r)))}}$ Sep 27 '18 at 11:46

It's easy to see that the denominator of $$\sum_{n=1}^i\frac1{n^r}$$ is at most $$i!^r$$, and the numerator is less than twice that for $$r>1$$. So we have a bound on the size of $$p+q$$ of order $$k^r$$. The prime number theorem suggests that, roughly, a $$\frac1{\log(k^r)}=\frac{1}{r\log k}$$ proportion of numbers about that size are prime. So if, instead of using your actual numbers, you just picked random numbers of about the right size, the chance of never hitting a prime would be $$\prod_{r=2}^{\infty}\big(1-\frac{1}{r\log k}\big)=0$$ (this equality follows from the divergence of the harmonic series).