Divisors

Note that we have the (loose, but good enough) bounds: $d_k=n$ ; $d_{k-1} \le \frac{n}{2}$ ; $d_{k-2} \le \frac{n}{3}$ ; $d_{k-i} \le \frac{n}{i+1}$ .

This means that the required sum satisfies the following inequality:

$\begin{aligned} d_1 \times d_2 + d_2 \times d_3 + \cdots + d_{k-1} \times d_k &\le \frac{n^2}{k \times (k-1)} + \frac{n^2}{(k-1) \times (k-2)} + \cdots + \frac{n^2}{2 \times 1} \\ &<n^2 \left(\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots \right) \\ &= n^2 \sum_{r=1}^{\infty} \frac{1}{r(r+1)}\\ &= n^2 \sum_{r=1}^{\infty} \left( \frac{1}{r} - \frac{1}{r+1} \right) \\ &=n^2 \end{aligned}$

since the final sum telescopes; hence the original sum is less than $n^2$ for all $n$ .