Bounds and limits

$p^{1/n}\to 1$: 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)

$(\genfrac{}{}{0ex}{}{a}{b})=\frac{a!}{b!(a-b)!}\approx \frac{a^a}{\sqrt{2n\pi }{b}^b(a-b{)}^{a-b}}$ for large a, b and a-b.

So: $(\genfrac{}{}{0ex}{}{a}{b})\le (\frac{ea}{b}{)}^b$. $\lceil \frac{a}{b}\rceil \le \frac{a-1}{b}+1$. $\frac{e^m}{\sqrt{em}}<\frac{m^m}{m!}\le \sum \frac{m^i}{i!}=e^m$. $(\frac{a}{b}{)}^b<(\genfrac{}{}{0ex}{}{a}{b})<a^b/b!$.

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 $x^2$: $e^{-x}\ge 1-x$.

$(1-x^{-1}{)}^x\le e^{-1}\le (1-x^{-1}{)}^{x-1}$: first inequality from taylor series argument. Latter ineq similarly proved for $x\in (0,1)$ by showing $e\ge (1-x^{-1}{)}^{1-x}$.

Also can get bounds using Taylor series; eg: $2^{-1}(e^{ac}+e^{-ac})\le e^{a^2{c}^2{2}^{-1}}$.

Bounds on log x

Using log series, or from concavity: $\ln x\le x-1$; 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. $a,b\ge 0$; $p,q\in R^{+}$ with $p+q=1$; then $ab\le p{a}^{\frac{1}{p}}+q{b}^{q^{-1}}$. 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 $a,b\ge 0$. Using the same technique (taking $\log (a^p{b}^q)$ and taking advantage of concavity of $\log (x)$), we see that weighted GM $\le$ weighted AM.

Also, weighted GM $\ge$ weighted HM: from applying the AM-GM inequality: $(a^p{b}^q{)}^{-1}\le p{a}^{-1}+q{b}^{-1}$.