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In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, then one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula one prove that the conjecture is true unconditionally.

In fact, such problem has been investigated by Erdös and Nathanson[On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 2}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by note that $p_n\sim n\log n$, one can also give the proof.

In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula, one can prove that the conjecture is true unconditionally.

In fact, this problem has been investigated by Erdős and Nathanson [On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 2}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by noting that $p_n\sim n\log n$, one can give an alternative proof.

In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, then one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula one prove that the conjecture is true unconditionally.

In fact, such problem has been investigated by Erdös and Nathanson[On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 1}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by note that $p_n\sim n\log n$, one can also give the proof.

In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, then one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula one prove that the conjecture is true unconditionally.

In fact, such problem has been investigated by Erdös and Nathanson[On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 2}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by note that $p_n\sim n\log n$, one can also give the proof.

In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, then one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log\log X}{\log X}. $$ Therefore, by using the standard Abel’s summation formula one prove that the conjecture is true unconditionally.

In fact, such problem has been investigated by Erdös and Nathanson[On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 1}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by note that $p_n\sim n\log n$, one can also give the proof.

In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that $$\sum_{n\le X}\frac{1}{p_{n+1}-p_n}\sim \frac{X\log\log X}{\log X},$$ and without the Hardy–Littlewood prime-pair conjecture, then one has $$ \sum_{n\le X}\frac{1}{p_{n+1}-p_n}\ll \frac{X\log\log X}{\log X}. $$ Therefore, by using Abel’s summation formula one prove that the conjecture is true unconditionally.

In fact, such problem has been investigated by Erdös and Nathanson[On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved $$ \sum_{n\ge 1}\frac{1}{(p_{n+1}-p_n)n(\log\log n)^c}<+\infty, $$ for all $c>2$. Then by note that $p_n\sim n\log n$, one can also give the proof.