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Henry
Posts:
1,089
Registered:
12/6/04
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Re: summing series
Posted:
Jan 21, 2011 8:29 PM
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On Jan 21, 10:50 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
> On Jan 21, 1:43 pm, maddy <madhuresh....@gmail.com> wrote:
>
> > how do we sum the series with terms a(n) = (1/8)^n (2n + 1)!/(n!)^2 to
> > infinity?
> > Thank you for your help.
>
> Maple 9.5 gets:
>
> an:=(1/8)^n *(2*n + 1)!/(n!)^2;
> n
> (1/8) (2 n + 1)!
> an := -----------------
> 2
> (n!)
>
> > S0:=sum(an,n=1..infinity);
>
> 1/2
> S0 := 2 2 - 1
> So the sum for n from 1 to infinity is 2*sqrt(2)-1 .
>
> We can check this numerically, by summing an for n from 1 to 50 and
> comparing:
> for k from 1 to 50 do t[k]:=evalf(subs(n=k,an)): end do:
> so t[1]=a1,t[2]=a2, etc. Note: values of an for n near 50 and above
> are very small: t[50] =~= 0.71396e(-14), so the sum for n from 51 to
> infinity is very small as well. We know this because the tail of the
> sum is much less than the corresponding geometric series. We get:
> sum_{n=1..50} an = 1.828427126, while 2*sqrt(2)-1 = 1.828427124 . The
> only mystery is how Maple gets the sum.
>
> R.G. Vickson
It might be more natural in this case to start from n=0 to give
sqrt(8) as the sum.
Indeed more generally
sum_{n=0 to infinity} x^n * (2n + 1)! / (n!)^2 = (1-4*x)^(-3/2),
at least for -1/4 < x < 1/4.
generatingfunctionology http://www.math.upenn.edu/~wilf/DownldGF.html
provides a variety of methods for questions like this.
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