Difference between r1.3 and the current
@@ -20,7 +20,12 @@
$1+\frac1{\sqrt{n}}>\sqrt[2n]{n}$
= tmp random links =
= wikiadmin =
RENAMETHISPAGE
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[[아벨_부등식,Abel_inequality]]?
{
https://mathworld.wolfram.com/AbelsInequality.html"Abel's inequality"
Ggl:"Abel's inequality"
}
= wikiadmin =
RENAMETHISPAGE
Ex 1 ¶
$\displaystyle (1+x)^n\ge 1+nx\quad\quad(n=1,2,\cdots)$
증명:
이항정리,binomial_theorem에서
이항정리,binomial_theorem에서
$\displaystyle (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots+x^n$
이고, $\displaystyle x>0$ 이기 때문에$\displaystyle (1+x)^n\ge 1+nx$
Ex 2 ¶
$\displaystyle 1+\frac1{\sqrt{n}}>\sqrt[2n]{n}\quad\quad(n=1,2,\cdots)$
증명
위 Ex 1 에서 $\displaystyle (1+x)^n\ge 1+nx$ 가 성립하고,
$\displaystyle \sqrt{n}>0$ 이므로, $\displaystyle x=\frac1{\sqrt{n}}$ 으로 두면
위 Ex 1 에서 $\displaystyle (1+x)^n\ge 1+nx$ 가 성립하고,
$\displaystyle \sqrt{n}>0$ 이므로, $\displaystyle x=\frac1{\sqrt{n}}$ 으로 두면
$\displaystyle \left(1+\frac1{\sqrt{n}}\right)^n\ge 1+\sqrt{n}>\sqrt{n}$
$\displaystyle 1+\frac1{\sqrt{n}}>\sqrt[2n]{n}$
$\displaystyle 1+\frac1{\sqrt{n}}>\sqrt[2n]{n}$
tmp random links ¶
아벨_부등식,Abel_inequality?
{
https://mathworld.wolfram.com/AbelsInequality.html
"Abel's inequality"
Abel's inequality
}
{
https://mathworld.wolfram.com/AbelsInequality.html
"Abel's inequality"
Abel's inequality
}