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Here I ask a question concerning segments of the divergent series $$\sum_p\frac1{p-1}=\sum_{k=1}^\infty\frac1{p_k-1},\tag{$*$}$$ where $p$ runs over all the primes, and $p_k$ denotes the $k$-th prime.

Question: Are all those segments $$\sum_{k=m}^n\frac1{p_k-1}\ \ (1\le m\le n)$$ of the series $(*)$ pairwise distinct?

On Sept. 9, 2015, I conjectured that the answer is positive and verified this for all $1\le m\le n\le2500$. Moreover, on the basis of my computation, I conjecture that if $d\in\{-1,0,1,2,\ldots\}$ and $$\sum_{k=m}^n\frac1{p_k+d}=\sum_{k=s}^t\frac1{p_k+d}\ \ \text{with}\ (m,n)\not=(s,t)\ \text{and}\ n\le t$$ then $d$ must be $1$, $(m,n)=(4,4)$ and $(s,t)=(8,10)$, or $(m,n)=(4,7)$ and $(s,t)=(5,10)$.

In contrast, P. Erdos and I. Niven [Bull. Amer. Math. Soc. 52(1946), 248-251] proved that all the segments $$\sum_{k=m}^n\frac1k\ \ (1\le m\le n)$$ of the harmonic series $\sum_{k=1}^\infty\frac1k$ are pairwise distinct.

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    $\begingroup$ Note that $$\frac1{p_4+1}=\frac1{p_8+1}+\frac1{p_9+1}+\frac1{p_{10}+1}=\frac18$$ and $$\sum_{k=4}^7\frac1{p_k+1}=\sum_{k=5}^{10}\frac1{p_k+1}=\frac{169}{504}.$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Nov 29, 2018 at 2:44
  • $\begingroup$ See also oeis.org/A261993. $\endgroup$
    – Zhi-Wei Sun
    Commented Nov 29, 2018 at 2:53

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