1.10. Basic inequalities 13 = 1 α α + 1 3 α3 + 1 5 α5 + · · · − 1 1 3 ( α2 + α4 + · · · ) = α2 3(1 − α2) = 1 12n(n + 1) , thus establishing that the sequence {an}n≥1 is decreasing and converges to some positive number c. Therefore, lim n→∞ ean = lim n→∞ n!en nn+ 1 2 = ec. It remains to show that (1.15) ec = √ 2π. To do so, we call upon the well-known Wallis formula for π (proved in 1656), namely lim n→∞ 2 · 4 · 6 · · · (2n) 1 · 3 · 5 · · · (2n − 1) √ 2n = π 2 , which can be written as (1.16) (2nn!)2 (2n)! 1 √ 2n ∼ π 2 as n → ∞. (For an elementary proof of this formula, see W¨ astlund [145].) Using the fact that n! ∼ nn √ ne−nec, we may replace n! by nn √ ne−nec in (1.16), to conclude after simplifications that ec ∼ √ 2π, which proves (1.15). Remark 1.15. Note that Stirling’s formula was actually proved by Abra- ham de Moivre (1667–1754, France) and later improved by James Stirling (1692–1770, Scotland), who established the value of the constant c. More- over, note that it is also possible to prove the following bounds: (1.17) 1 ≤ n! nne−n √ 2πn ≤ e1/12n (n ≥ 1) (see Problem 1.15). 1.10. Basic inequalities Two very useful inequalities are the Arithmetic Geometric Mean inequality (AGM inequality for short) and the Cauchy-Schwarz inequality, which we state respectively as follows.

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