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[SPAM] [obm-l] Fw: [obm-l] Fw: [obm-l] soma de série

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Subject: [SPAM] [obm-l] Fw: [obm-l] Fw: [obm-l] soma de série
From: Luiz Alberto Duran Salomão <salomao@xxxxxx>
Date: Tue, 11 Mar 2008 18:01:12 -0300
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This is a multi-part message in MIME format.

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Caro Saulo,
No seu exemplo, o mmc de 2, 3 e 4 =E9 12 (e n=E3o 24).
Portanto, o que temos =E9
1+1/2+1/3+1/4 =3D (12+6+4+3)/12 =3D 25/12, ou seja, =EDmpar/par.
Abra=E7os,
Luiz Alberto
----- Original Message -----=20
From: saulo nilson=20
To: obm-l@xxxxxxxxxxxxxx=20
Sent: Tuesday, March 11, 2008 3:32 PM
Subject: Re: [obm-l] Fw: [obm-l] soma de s=E9rie


1+1/2+1/3+1/4=3D(24+12+8+6)/24=3Dpar /par


2008/3/11 Luiz Alberto Duran Salom=E3o <salomao@xxxxxx>:


  ----- Original Message -----=20
  From: Luiz Alberto Duran Salom=E3o=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Tuesday, March 11, 2008 10:21 AM
  Subject: Re: [obm-l] soma de s=E9rie


  Caros amigos,
  Seja n um inteiro, com n>1. O que se quer provar =E9 que
  1+1/2+1/3+ . . . +1/n   n=E3o =E9 inteiro.
  Seja 2^a a maior pot=EAncia de 2 tal que 2^a =E9 menor do que ou igual =
a n.
  Assim, 1/2^a   aparece no somat=F3rio acima mas 1/2^(a+1)  n=E3o =
aparece.
  Observe que o m=EDnimo m=FAltiplo comum dos denominadores dos=20
  termos do somat=F3rio tem a pot=EAncia 2^a como fator. Agora, no=20
  numerador de cada  fra=E7=E3o, j=E1 com denominador igual ao m=EDnimo
  m=FAltiplo comum, temos sempre um n=FAmero par, com exce=E7=E3o do=20
  termo correspondente a 1/2^a. Logo, a soma dos numeradores=20
  =E9 =EDmpar o que nos leva a concluir que o somat=F3rio n=E3o =E9  um=20
  n=FAmero inteiro.
  Abra=E7os,
  Luiz Alberto
    ----- Original Message -----=20
    From: MauZ=20
    To: obm-l@xxxxxxxxxxxxxx=20
    Sent: Monday, March 10, 2008 6:13 PM
    Subject: [obm-l] soma de s=E9rie


    mostrar que 1+1/2+1/3+...+1/n n=E3o =E9 inteiro pra qquer N>1.

    Obrigado!



-------------------------------------------------------------------------=
---


    No virus found in this incoming message.
    Checked by AVG.=20
    Version: 7.5.518 / Virus Database: 269.21.7/1324 - Release Date: =
10/3/2008 19:27







-------------------------------------------------------------------------=
-------


No virus found in this incoming message.
Checked by AVG.=20
Version: 7.5.518 / Virus Database: 269.21.7/1324 - Release Date: =
10/3/2008 19:27

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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2900.3268" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Caro Saulo,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>No seu exemplo, o mmc de 2, 3 e 4 =E9 =
12 (e n=E3o=20
24).</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Portanto, o que temos =E9</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>1+1/2+1/3+1/4&nbsp;=3D (12+6+4+3)/12 =
=3D 25/12, ou=20
seja, =EDmpar/par.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Abra=E7os,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Luiz Alberto</FONT></DIV>
<DIV style=3D"FONT: 10pt arial">----- Original Message -----=20
<DIV style=3D"BACKGROUND: #e4e4e4; font-color: black"><B>From:</B> <A=20
title=3Dsaulo.nilson@xxxxxxxxx =
href=3D"mailto:saulo.nilson@xxxxxxxxx";>saulo=20
nilson</A> </DIV>
<DIV><B>To:</B> <A title=3Dobm-l@xxxxxxxxxxxxxx=20
href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
<DIV><B>Sent:</B> Tuesday, March 11, 2008 3:32 PM</DIV>
<DIV><B>Subject:</B> Re: [obm-l] Fw: [obm-l] soma de s=E9rie</DIV></DIV>
<DIV><BR></DIV>1+1/2+1/3+1/4=3D(24+12+8+6)/24=3Dpar /par<BR><BR>
<DIV class=3Dgmail_quote>2008/3/11 Luiz Alberto Duran Salom=E3o &lt;<A=20
href=3D"mailto:salomao@xxxxxx";>salomao@xxxxxx</A>&gt;:<BR>
<BLOCKQUOTE class=3Dgmail_quote=20
style=3D"PADDING-LEFT: 1ex; MARGIN: 0px 0px 0px 0.8ex; BORDER-LEFT: #ccc =
1px solid">
  <DIV bgcolor=3D"#ffffff">
  <DIV class=3DIh2E3d>
  <DIV>&nbsp;</DIV>
  <DIV style=3D"FONT: 10pt arial">----- Original Message -----=20
  <DIV style=3D"BACKGROUND: #e4e4e4"><B>From:</B> <A =
title=3Dsalomao@xxxxxx=20
  href=3D"mailto:salomao@xxxxxx"; target=3D_blank>Luiz Alberto Duran =
Salom=E3o</A>=20
  </DIV>
  <DIV><B>To:</B> <A title=3Dobm-l@xxxxxxxxxxxxxx=20
  href=3D"mailto:obm-l@xxxxxxxxxxxxxx"; =
target=3D_blank>obm-l@xxxxxxxxxxxxxx</A>=20
  </DIV>
  <DIV><B>Sent:</B> Tuesday, March 11, 2008 10:21 AM</DIV>
  <DIV><B>Subject:</B> Re: [obm-l] soma de s=E9rie</DIV></DIV>
  <DIV><BR></DIV>
  <DIV><FONT face=3DArial size=3D2>Caros amigos,</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Seja n um inteiro, com n&gt;1. O que =
se quer=20
  provar =E9 que</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>1+1/2+1/3+ . . . +1/n&nbsp;&nbsp; =
n=E3o =E9=20
  inteiro.</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Seja 2^a a maior pot=EAncia de 2 tal =
que 2^a =E9=20
  menor do que ou igual a n.</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Assim, 1/2^a&nbsp;&nbsp; aparece no =
somat=F3rio=20
  acima mas 1/2^(a+1)&nbsp; n=E3o aparece.</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Observe que o m=EDnimo m=FAltiplo =
comum dos=20
  denominadores dos </FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>termos do somat=F3rio tem a =
pot=EAncia 2^a como=20
  fator. Agora, no </FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>numerador de cada&nbsp; fra=E7=E3o, =
j=E1 com=20
  denominador igual ao m=EDnimo</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>m=FAltiplo comum, temos sempre um =
n=FAmero par, com=20
  exce=E7=E3o do </FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>termo correspondente a 1/2^a. Logo, a =
soma dos=20
  numeradores </FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>=E9 =EDmpar o que nos leva a concluir =
que o somat=F3rio=20
  n=E3o =E9&nbsp; um </FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>n=FAmero inteiro.</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Abra=E7os,</FONT></DIV>
  <DIV><FONT face=3DArial size=3D2>Luiz Alberto</FONT></DIV></DIV>
  <BLOCKQUOTE=20
  style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
    <DIV class=3DIh2E3d>
    <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
    <DIV style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial"><B>From:</B> <A =

    title=3Dmauz.matematica@xxxxxxxxx =
href=3D"mailto:mauz.matematica@xxxxxxxxx"=20
    target=3D_blank>MauZ</A> </DIV>
    <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
    href=3D"mailto:obm-l@xxxxxxxxxxxxxx"; =
target=3D_blank>obm-l@xxxxxxxxxxxxxx</A>=20
    </DIV>
    <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Monday, March 10, 2008 =
6:13=20
    PM</DIV></DIV>
    <DIV class=3DIh2E3d>
    <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] soma de =
s=E9rie</DIV>
    <DIV><BR></DIV>mostrar que 1+1/2+1/3+...+1/n n=E3o =E9 inteiro pra =
qquer=20
    N&gt;1.<BR><BR>Obrigado!<BR></DIV>
    <DIV class=3DIh2E3d>
    <P>
    <HR>

    <P></P>No virus found in this incoming message.<BR>Checked by AVG.=20
    <BR>Version: 7.5.518 / Virus Database: 269.21.7/1324 - Release Date: =

    10/3/2008 19:27<BR>
    <P></P>
    <P></P></DIV></BLOCKQUOTE></DIV></BLOCKQUOTE></DIV><BR>
<P>
<HR>

<P></P>No virus found in this incoming message.<BR>Checked by AVG. =
<BR>Version:=20
7.5.518 / Virus Database: 269.21.7/1324 - Release Date: 10/3/2008=20
19:27<BR></BODY></HTML>

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