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.

For any $n \ge 8$, we have $$\dfrac{n-2}{\binom{n}{2}} = \dfrac{n-2}{\tfrac{n(n-1)}{2}} = \dfrac{2}{n} \cdot \dfrac{n-2}{n-1} \le \dfrac{2}{n} \le \dfrac{1}{4},$$
and thus, $$\sum_{k = 3}^{n-1}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n-1}{4} \implies \sum_{k = 3}^{n}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n}{4}.$$
So if you can prove that $\displaystyle\sum_{k = 3}^{n}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n}{4}$ for $3 \le n \le 7$, then you can use induction to finish the proof.
Remark: Obviously, Kevin Song's analysis gives a much better asymptotic bound. But if you only need to prove the upper bound $\dfrac{n}{4}$, this might be simpler.
