#1
|
||||
|
||||
͹ءÃÁ P
ÍÂÒ¡·ÃÒºÇèÒ·ÓäÁ ͹ءÃÁ $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{n}$ à»ç¹Í¹Ø¡ÃÁä´àÇÍÃìਹ·ì ÁÕÇÔ¸Õ¾ÔÊÙ¨¹ìÁÑê¤ÃѺ
áÅéÇ·ÓäÁ $\frac{1}{1^m}+\frac{1}{2^m}+\frac{1}{3^m}+...+\frac{1}{n^m}$ àÁ×èÍ $ m > 1$à»ç¹Í¹Ø¡ÃÁ¤Í¹àÇÍÃìਹ·ì -*- ¢ÍÇÔ¸Õ¾ÔÊÙ¨¹ì´éÇ áÅéÇ¡ç½Ò¡â¨·Âì¢é͹Õé´éǤÃѺ $\frac{\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+...}{\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+...} = ??$
__________________
àÁ×èÍäÃàÃÒ¨Ðà¡è§àÅ¢¹éÒÒÒÒÒÒ ~~~~ T T äÁèà¡è§«Ñ¡·Õ ·Óä§´Õ |
#2
|
||||
|
||||
ÍéÒ§ÍÔ§:
$\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{6^3}+\frac{1}{8^3}+...=\frac{n}{8}$ ¨Ðä´é $\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+...=n-2(\frac{n}{8} )$ áÅÐ$\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+...=n-\frac{n}{8} $ $\frac{\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+...}{\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+...} = \frac{n-2(\frac{n}{8} )}{n-\frac{n}{8}}=\frac{6}{7} $
__________________
à˹×Í¿éÒÂѧÁÕ¿éÒáµèà˹×Í¢éÒµéͧäÁèÁÕã¤Ã »Õ¡¢Õé¼×駢ͧ»ÅÍÁ§Ñé¹ÊԹР...âÅ¡¹ÕéâË´ÃéÒ¨ÃÔ§æ ÁѹãËé¤ÇÒÁÊØ¢¡ÑºàÃÒ áÅéÇÊØ´·éÒ Áѹ¡çàÍҤ׹ä»... |
#3
|
||||
|
||||
¢éÍ 1 ¼ÁÇèÒà»ç¹¤Í¹àÇÍÃìਹ·ì¹Ð¤ÃѺ
¢éÍ 2 ãËé $S = 1+\frac{1}{2^m}+\frac{1}{3^m}+...$ ¹èҨоÔÊÙ¨¹ìÇèÒ $\frac{1}{n^m}>\frac{1}{(n+1)^m}$ ·Ø¡ $n\in N$ áÅÐ $m>1$ ¢éÍ 3 ä´é $\frac{6}{7}$ à·èҤس light ¤ÃѺ ÁÕËÅÒÂÇÔ¸Õ¤Ô´¹Ð¤ÃѺ¢é͹Õé = $.\frac{\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+...}{\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3+...}}-1+1$ = $.\frac{-\frac{1}{2^3}-\frac{1}{4^3}-\frac{1}{6^3}-...}{\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3+...}}+1$ = $.(\frac{\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+...}{-\frac{1}{2^3}(\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+...)})^{-1}+1$ = $.\frac{6}{7}$ 10 ÁÕ¹Ò¤Á 2009 01:21 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ [SIL] à˵ؼÅ: â¤é´á»ê¡ |
#4
|
|||
|
|||
ÍéÒ§ÍÔ§:
ÊÓËÃѺ ¡Ã³Õ $p=1$ ÊÒÁÒö¾ÔÊÙ¨¹ìâ´Â¡ÒÃãªéÍÊÁ¡Òáçä´é ÊѧࡵÇèÒ $1+\dfrac{1}{2}+\cdots+\dfrac{1}{2^n}=1+\dfrac{1}{2}+\Big(\dfrac{1}{3}+\dfrac{1}{4}\Big)+\Big(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1 }{7}+\dfrac{1}{8}\Big)+\cdots+\Big(\dfrac{1}{2^{n-1}+1}+\cdots+\dfrac{1}{2^n}\Big)$ $~~~~~~~~~~~~~~~~~~~~~~>1+\dfrac{1}{2}+\Big(\dfrac{1}{4}+\dfrac{1}{4}\Big)+\Big(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1} {8}\Big)+\cdots+\Big(\dfrac{1}{2^n}+\cdots+\dfrac{1}{2^n}\Big)$ $~~~~~~~~~~~~~~~~~~~~~~=1+\underbrace{\dfrac{1}{2}+\cdots+\dfrac{1}{2}}_{n}$ $~~~~~~~~~~~~~~~~~~~~~~=1+\dfrac{n}{2}$ ãËé $n\to\infty$ ¨Ðä´éÇèÒ $1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots = \infty$
__________________
site:mathcenter.net ¤Ó¤é¹ |
#5
|
||||
|
||||
ÅÙèÍÍ¡ËÃÍà¹Õ蠢ͺ¤Ø³ÊÓËÃѺÇÔ¸Õ¤Ô´¤ÃѺ
|
|
|