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Topic:
summing series
Replies:
7
Last Post:
Jan 22, 2011 1:32 PM



Henry
Posts:
1,089
Registered:
12/6/04


Re: summing series
Posted:
Jan 21, 2011 8:29 PM


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 = (14*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.



