<s>[INST] How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, however I can't find a way to prove it.
 [/INST] $\displaystyle \sum\limits_{i=3}^n (i-2){\binom i 2}^{-1} = 2\sum\limits_{i=3}^n \frac{i-2}{i(i-1)}=2H_{n-1}-4+\frac{4}{n}$
From the assumption $\displaystyle~2H_{n-1}-4+\frac{4}{n}<\frac{n}{4}~$ we conclude for $~n\to n+1~$ :
$\displaystyle 2H_n-4+\frac{4}{n+1}= \left(2H_{n-1}-4+\frac{4}{n}\right) + \left(\frac{4}{n+1}-\frac{2}{n}\right)< $
$\displaystyle < \frac{n}{4} + \frac{4}{n+1}-\frac{2}{n}<\frac{n+1}{4}$
The right side of the inequation chain means $~n^2-7n+8>0~$, 
it's the same as $\displaystyle~n>4+\frac{n}{n-2}~$, which is true for $~n\geq 6~$ .
The cases $~n\in\{3,4,5\}~$ can be checked by hand.
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