<s>[INST] Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to show it.
 [/INST] The sum can rewrite in this way:
$$\frac{1}{(n-2)(n-1)n}\sum_{r=1}^n(n^2-nr-n-nr+r^2+r)=$$
$$=\frac{1}{(n-2)(n-1)n}\left(n^3-n(\frac{n(n+1)}{2})-n^2-n(\frac{n(n+1)}{2})+(\frac{n(n+1)(2n+1)}{6})+(\frac{n(n+1)}{2})\right)=$$
$$=\frac{1}{6(n-2)(n-1)n}(6n^3-3n^3-3n^2-6n^2-3n^3-3n^2+2n^3+3n^2+n+3n^2+3n)=$$
$$=\frac{1}{6(n-2)(n-1)n}(2n^3-6n^2+4n)=\frac{1}{6(n-2)(n-1)}(2n^2-6n+4)=$$
$$=\frac{1}{3(n-2)(n-1)}(n^2-3n+2)=\frac{(n-2)(n-1)}{3(n-2)(n-1)}=\frac{1}{3}$$
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