\(p^{1/n} \to 1\): take seq \(x_{n} = p^{1/n}-1\). Also, \(n^{1/n} \to 1\).
Bounds on combinatorial quantities
Bounds on n!
(Stirling)\(n!\approx \sqrt{2n\pi} (\frac{n}{e})^{n}\). \why
Bounds on (a choose b)
\(\binom{a}{b} = \frac{a!}{b!(a-b)!} \approx \frac{a^{a}}{\sqrt{2n\pi} b^{b}(a-b)^{a-b}}\) for large a, b and a-b.
So: \(\binom{a}{b} \leq (\frac{ea}{b})^{b}\). \(\lceil \frac{a}{b}\rceil \leq \frac{a-1}{b}+1\). \(\frac{e^{m}}{\sqrt{em}}< \frac{m^{m}}{m!}\leq \sum \frac{m^{i}}{i!} = e^{m}\). \((\frac{a}{b})^{b} < \binom{a}{b} < a^{b}/b!\).
Bounds on powers and exponents
Bounds on \htext{\((1+b)^{n\) and \(e^{n}\)}{exponentials}}
Using Taylor series of \(e^{-x}\) wrt 0, stopped at \(x^{2}\): \(e^{-x} \geq 1-x\).
\((1-x^{-1})^{x} \leq e^{-1} \leq (1-x^{-1})^{x-1}\): first inequality from taylor series argument. Latter ineq similarly proved for \(x \in (0, 1)\) by showing \(e \geq (1-x^{-1})^{1-x}\).
Also can get bounds using Taylor series; eg: \(2^{-1}(e^{ac}+e^{-ac}) \leq e^{a^{2}c^{2}2^{-1}}\).
Bounds on log x
Using log series, or from concavity: \(\ln x \leq x-1\); easy to visualize with log x curve under x - 1 line.
Inequalities among means
Bound ab with a convex combination of powers
Aka Young’s inequality. \(a, b \geq 0\); \(p, q \in R^{+}\) with \(p + q = 1\); then \(ab \leq pa^{\frac{1}{p}} + qb^{q^{-1}}\). Special case of the inequality of weighted arithmetic and geometric means, but is described separately to illustrate the proof technique.
Pf: Let \(a, b >0\). \(\ln ab = \ln a + \ln b = p\ln a^{p^{-1}} + q\ln b^{q^{-1}}\). \(\ln(x)\) is strictly concave function on \(R^{+}\); thence use this to get inequality.
AM GM HM inequality
The various means are described in the statistics survey.
Suppose \(a, b \geq 0\). Using the same technique (taking \(\log(a^{p}b^{q})\) and taking advantage of concavity of \(\log(x)\)), we see that weighted GM \(\leq\) weighted AM.
Also, weighted GM \(\geq\) weighted HM: from applying the AM-GM inequality: \((a^p b^q)^{-1} \leq pa^{-1} + qb^{-1}\).