$Ln(2)=0$
Inviato: 07/10/2015, 00:02
Dato che ...
$ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6+...$
allora ...
$ln(2)=1-1/2+1/3-1/4+1/5-1/6+...$
$ln(2)=(1+1/3+1/5+...)-(1/2+1/4+1/6+...)$
$ln(2)=[(1+1/3+1/5+...)+(1/2+1/4+1/6+...)]-2(1/2+1/4+1/6+...)$
$ln(2)=[1+1/2+1/3+1/4+1/5+1/6+...]-(1+1/2+1/3+1/4+1/5+1/6+...)$
$ln(2)=0$
... what?
Cordialmente, Alex
$ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6+...$
allora ...
$ln(2)=1-1/2+1/3-1/4+1/5-1/6+...$
$ln(2)=(1+1/3+1/5+...)-(1/2+1/4+1/6+...)$
$ln(2)=[(1+1/3+1/5+...)+(1/2+1/4+1/6+...)]-2(1/2+1/4+1/6+...)$
$ln(2)=[1+1/2+1/3+1/4+1/5+1/6+...]-(1+1/2+1/3+1/4+1/5+1/6+...)$
$ln(2)=0$
... what?
Cordialmente, Alex