summing series
maddy
infinity?
Thank you for your help.
sanebow
"maddy" <madhur...@gmail.com> wrote in message
news:88b672cc-b573-4c6d...@t8g2000prh.googlegroups.com...
> how do we sum the series with terms a(n) = (1/8)^n (2n + 1)!/(n!)^2 to
> infinity?
> Thank you for your help.
a(n)/a(n+1) = 8* (1 / ( ( 2*n+3 ) * ( 2*n+2 ) ) ) * 1/(n+1)
series diminishes by about 1/(n^3) ( fast)
Ray Vickson
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
Henry
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.
Henry
It might be more natural in this case to start from n=0 to give
Rob Johnson
Consider the sum
oo
--- n
f(x) = > (2n+1) C(2n,n) x [1]
---
k=0
Your sum is f(1/8). This is reminiscent of the series
oo
-1/2 --- n
(1-4x) = > C(2n,n) x [2]
---
n=0
Taking the derivative, we get
oo
-3/2 --- n-1
2 (1-4x) = > n C(2n,n) x [3]
---
n=0
Adding [2] and 2x times [3] we get
-3/2
f(x) = (1-4x) [4]
Thus, your sum is f(1/8) = 2^{-3/2}.
Rob Johnson <r...@trash.whim.org>
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Rob Johnson
Ray Vickson
Of course, he says (striking his forehead as he does so). Had I put x
instead of 1/8 in the sum, Maple would have given me this. Of course,
once we see the answer we can easily derive it, just using the
expression for the "negative binomial" C(-a,n) with a > 0.
R.G. Vickson
>
> generatingfunctionologyhttp://www.math.upenn.edu/~wilf/DownldGF.html