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....
I would glad you guys to think about this thing and share opinions for me and my friend's curiosity.
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.)
It's likely to be very difficult to prove (or disprove) but I would guess it was true. Here is a heuristic for why I think this.
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).
Of course, your numbers aren't random and aren't independent of each other, but in the absence of a good reason why your process is less likely to eventually produce a prime, I suspect it will.