Return to Answer

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below).

It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

Added. It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$.

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below).

It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

Added. It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$. As Lucia kindly pointed out, a result of Gallagher's implies that $C=1$.

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below).

It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

Added. It seems that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$.

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below).

It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

Added. It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$.

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures. It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below). 

It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.

Added. It seems that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$.