prove that [1(2)+..... (elementary algebra)

Question

Prove {[1(2)+2(3)+....+n(n+1)]/n(n+3)} ≥ {[n+1]/4} for n≥1...

Answer 1

The general term is n(n+1)=n^2+n.

The sum is therefore the sum of the natural numbers up to n=n(n+1)/2 plus the sum of their squares=n^3/3+n^2/2+n/6. Therefore combined, these are n^2/2+n/2+n^3/3+n^2/2+n/6=n^3/3+n^2+2n/3.

This can be written n(n^2+3n+2)/3=n(n+1)(n+2)/3=S, where S is the sum of the series.

S/(n(n+3))=(n+1)(n+2)/(3(n+3)). What is the value of (n+2)/(3(n+3))? The minimum value is when n=1, (3)/(3(4))=1/4. So S/(n(n+3)) is less than or greater than (n+1)/4. When n>>1 (very much greater than 1), this expression gets closer to 1/3 and 1/3>1/4, so S/(n(n+3))>(n+1)/4.