MathLabs

Problem 2

Let m,nm,n be positive integers with n≤mn\le m. Prove that 2nn!≤(m+n)!(m−n)!≤(m2+m)n2^n n!\le\dfrac{(m+n)!}{(m-n)!}\le(m^2+m)^n.
Step 3 of 4: Establish the termwise bounds
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2i≤m2+m−i2+i≤m2+m(1≤i≤n≤m)2i\le m^2+m-i^2+i\le m^2+m\qquad(1\le i\le n\le m)
Detailed analysis

For the upper bound, −i2+i≤0-i^2+i\le0 because i≥1i\ge1. For the lower bound, m2+m−i2+i≥2im^2+m-i^2+i\ge2i is equivalent to m2+m≥i2+im^2+m\ge i^2+i, true because i≤mi\le m and u2+uu^2+u is increasing for positive uu.