I am trying to show that if the arithmetic mean of the products of all distinct pairs of positive integers whose sum is $n$ is denoted by $S_{n}$ then $$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \dfrac{1}{6}$$ Solution attempt
If $n$ was even then $$S_{n} = \sum _{i=1}^{i=\dfrac{n} {2}}\dfrac {\left( n-i\right) \left( n-\left( n-i\right) \right) } {\dfrac {n } {2}} = \sum _{i=1}^{i=\dfrac{n} {2}}2i\left( 1-\dfrac {i} {n}\right) $$
so $$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \lim _{n\rightarrow \infty } \sum _{i=1}^{i=\dfrac{n} {2}}\dfrac{2i}{n^{2}}\left( 1-\dfrac {i} {n}\right)$$
Similarly If $n$ was odd then $$S_{n} = \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}\dfrac {\left( n-i\right) \left( n-\left( n-i\right) \right) } {\dfrac {\left(n+1\right) } {2}} = \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}2i\left(\dfrac {n-i} {n+1}\right)$$so $$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \lim _{n\rightarrow \infty } \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}\dfrac{2i}{n^{2}}\left(\dfrac {n-i} {n+1}\right)$$
I am unsure how to proceed from here to show the result. Any help would be much appreciated.
