I have the following equations: $$q_b^{(n)}=\frac{\sum_\limits{k=2}^n{p_{k,b}^{(n)}kE(T_k^{(n)})}}{\sum_\limits{k=2}^n{kE(T_k^{(n)})}}$$ \ $$p_{k,b}^{(n)}=\frac{{{n-b-1}\choose{k-2}}}{{{n-1}\choose{k-1}}}$$ \ $$\sum_\limits{k=2}^n{{{n-k}\choose{b-1}}}={{n-1}\choose{b}}$$ \ $$E({T_k^{(n)}})=\frac{2}{k(k-1)}$$ \ I want to show that the following is true: $$q_b^{(n)}=\frac{\frac{1}{b}}{\sum_\limits{k=1}^{n-1}{\frac{1}{k}}}\propto\frac{1}{b}$$ \ Substituting into my original equation, I get $$q_b^{(n)}=\frac{\sum_\limits{k=2}^n\frac{{{n-b-1}\choose{k-2}}}{{{n-1}\choose{k-1}}}k\frac{2}{k(k-1)}}{\sum_\limits{k=2}^n{k\frac{2}{k(k-1)}}}$$ \ This can be simplified a bit to: $$q_b^{(n)}=\frac{2\sum_\limits{k=2}^n\frac{{{n-b-1}\choose{k-2}}}{{{n-1}\choose{k-1}}}\frac{1}{(k-1)}}{2\sum_\limits{k=2}^n{\frac{1}{(k-1)}}}$$ \ This can be cleaned up further to read: $$q_b^{(n)}=\frac{\sum_\limits{k=2}^n\frac{{{n-b-1}\choose{k-2}}}{{{n-1}\choose{k-1}}}\frac{1}{(k-1)}}{\sum_\limits{k=1}^{n-1}{\frac{1}{(k)}}}$$ \ Now I just need to show that $${\sum_\limits{k=2}^n\frac{{{n-b-1}\choose{k-2}}}{{{n-1}\choose{k-1}}}\frac{1}{(k-1)}}=\frac{1}{b}$$ \ Expanding the denominator I get $${\sum_{k=2}^n\frac{{{n-b-1}\choose{k-2}}}{\frac{(n-1)!}{(k-1)!(n-k-2)!}}\frac{1}{(k-1)}}$$
This is equal to $${\sum_{k=2}^n\frac{{{n-b-1}\choose{k-2}}(k-2)!(n-k-2)!}{{(n-1)!}}}$$
Expanding the binomial coefficient gives me $${\sum_\limits{k=2}^n\frac{(n-b-1)!(k-2)!(n-k-2)!}{{(n-1)!(k-2)!(n-b-k-3)!}}}$$
Cancelling the $(k-2)!$ gives me $${\sum_\limits{k=2}^n\frac{(n-b-1)!(n-k-2)!}{{(n-1)!(n-b-k-3)!}}}$$
But now I'm not sure where to go from here to get this to look like $1/b$