Re: A89: Re: help me too please!
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Re: A89: Re: help me too please!
On Fri, 21 May 1999 ZeromusMog@aol.com wrote:
> > (-1)^n/(2n+1) from n=0 to infinity. As each progressive addition
> > operation is performed, it gets closer and closer to pi. Yes, it will
> > never reach pi, and this is immediately obvious if you consider that it
> > is producing a rational number each time.
>
> I did this:
> 0->n:0->y:Lbl a:((-1)^n/(2n+1))+y->y:n+1->n:Disp y:Goto a
> Then pressed <>+Enter, and it wasn't getting closer to Pi... It went to
> 1.78....
It did? Interesting, it doesn't here.
The series given approximates arctan(1), which is pi/4. If you OTOH take
4*sigma((-1)^n/(2n+1),n,0,x) where x is an appropriately large value you
will approximate pi.
--
naisbodo@naisbodo.com
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