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.

The inequality is equivalent to:
$$\sum_{i=3}^{n} \frac{i-2}{i(i-1)} < \frac n 8$$
Using partial fractions, $\frac{i-2}{i(i-1)} = \frac 2 n - \frac 1 {n-1} = \frac 1 n - \left(\frac 1 {n-1} - \frac 1 n\right)$.
The given sum becomes:
$$\sum_{i=3}^{n} \frac 1 i - \sum_{i=3}^{n} \left(\frac 1{i-1} - \frac 1 i\right)$$
$$= H_n - \frac 3 2 - \left( \frac 1 2 - \frac 1 3 + \frac 1 3 - \frac 1 4 + \dots + \frac 1 {n-1} - \frac 1 n \right)$$
$$= H_{n} + \frac 1 n - 2$$
Where $H_n$ is the $n$th harmonic number.
The problem then transforms to proving the following inequality:
$$H_{n} + \frac 1 n - 2 < \frac{n}{8}, \qquad n \ge 3$$
which can be done by checking cases for $n < 8$ and induction for $n \ge 8$.
