MathLabs

Problem 4

Let n≥2n\ge2 be a positive integer with divisors 1=d1<d2<⋯<dk=n1=d_1<d_2<\cdots<d_k=n. Prove that d1d2+d2d3+⋯+dk−1dkd_1d_2+d_2d_3+\cdots+d_{k-1}d_k is always less than n2n^2, and determine when it is a divisor of n2n^2.
Step 2 of 3: Prove the strict bound
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In plain words

Prove the strict bound

∑i=1k−11didi+1≤∑i=1k−11i(i+1)<1\sum_{i=1}^{k-1}\frac1{d_id_{i+1}}\le\sum_{i=1}^{k-1}\frac1{i(i+1)}<1
Detailed analysis

Because di≥id_i\ge i, the reciprocal sum is at most ∑(1/i−1/(i+1))=1−1/k<1\sum(1/i-1/(i+1))=1-1/k<1. Hence S<n2S<n^2.