A curious limit
Rotwang
Does anybody know how to evaluate the limit as x -> 0 of
x^2 * Sum_{n = 1}^oo [n*exp(-n*x)/(1-exp(-2n*x))] ?
For reasons I won't go into (it's physics-related) it would be very
nice if the answer turned out to be pi^2/8, and based on numerical
evidence this appears to be true. But I have no idea how to prove it.
victor_me...@yahoo.co.uk
Let q = exp(-x). Then
sum_{n=1}^infinity n q^n/(1-q^{2n})
= sum_{n=1}^infinity sum_{m=1}^infinity n q^{n(2m-1)}
= sum_{m=1}^infinity sum_{n=1}^infinity n q^{n(2m-1)}
= sum_{m=1}^infinity q^{2m-1}/(1 - q^{2m-1})^2.
As
x^2 q^{2m-1}/(1 - q^{2m-1})^2
= x^2 exp(-(2m-1)x)/(1 - exp(-(2m-1)x))^2 -> 1/(2m-1)^2
as x -> 0 the original limit is
sum_{m=1}^infinity 1/(2m-1)^2 = pi^2/8.
Victor Meldrew
"I don't believe it!"
Rotwang
> On 5 Nov, 18:21, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > Does anybody know how to evaluate the limit as x -> 0 of
>
> > x^2 * Sum_{n = 1}^oo [n*exp(-n*x)/(1-exp(-2n*x))] ?
>
> > For reasons I won't go into (it's physics-related) it would be very
> > nice if the answer turned out to be pi^2/8, and based on numerical
> > evidence this appears to be true. But I have no idea how to prove it.
>
> Let q = exp(-x). Then
> sum_{n=1}^infinity n q^n/(1-q^{2n})
> = sum_{n=1}^infinity sum_{m=1}^infinity n q^{n(2m-1)}
> = sum_{m=1}^infinity sum_{n=1}^infinity n q^{n(2m-1)}
> = sum_{m=1}^infinity q^{2m-1}/(1 - q^{2m-1})^2.
OK...
> As
> x^2 q^{2m-1}/(1 - q^{2m-1})^2
> = x^2 exp(-(2m-1)x)/(1 - exp(-(2m-1)x))^2 -> 1/(2m-1)^2
> as x -> 0 the original limit is
> sum_{m=1}^infinity 1/(2m-1)^2 = pi^2/8.
Very nice, thanks. One step I'm not sure about though - how do you
show that the sum over m commutes with the limit x->0 in this case?
W^3
<de072f85-325d-41eb...@s1g2000prg.googlegroups.com>,
Rotwang <sg...@hotmail.co.uk> wrote:
Let h = -(2m-1)x and write the mth term as [h^2*e^h/(1 -
e^h)^2]*(1/(2m-1)^2). Then show the term is brackets is bounded on R.
You can then use the Weierstrass M test to see the sum is uniformly
convergent for x in a neighborhood of 0. This allows interchange of
limit and sum.
Rotwang
> In article
> <de072f85-325d-41eb-a438-19012c740...@s1g2000prg.googlegroups.com>,
Thanks. I will probably be using this result in my PhD thesis - do you
have a real name so I can acknowledge you? The same question goes to
Victor.
W^3
<c7f28bc9-2b8d-4c0c...@1g2000prd.googlegroups.com>,
Rotwang <sg...@hotmail.co.uk> wrote:
How about "W^3 from sci.math"?
W^3
<c7f28bc9-2b8d-4c0c...@1g2000prd.googlegroups.com>,
Rotwang <sg...@hotmail.co.uk> wrote:
By the way, the above is better stated as: Show that h^2*e^h/(1 -
e^h)^2 is bounded on R. This implies there exists C such that the mth
term is bounded by C/(2m-1)^2, m = 1, 2, ... etc.
Rotwang
> Rotwang <sg...@hotmail.co.uk> wrote:
>
> [...]
>
> > Thanks. I will probably be using this result in my PhD thesis - do you
> > have a real name so I can acknowledge you? The same question goes to
> > Victor.
>
> How about "W^3 from sci.math"?
Sure.
victor_me...@yahoo.co.uk
>
> > Let q = exp(-x). Then
> > sum_{n=1}^infinity n q^n/(1-q^{2n})
> > = sum_{n=1}^infinity sum_{m=1}^infinity n q^{n(2m-1)}
> > = sum_{m=1}^infinity sum_{n=1}^infinity n q^{n(2m-1)}
> > = sum_{m=1}^infinity q^{2m-1}/(1 - q^{2m-1})^2.
>
> OK...
>
> > As
> > x^2 q^{2m-1}/(1 - q^{2m-1})^2
> > = x^2 exp(-(2m-1)x)/(1 - exp(-(2m-1)x))^2 -> 1/(2m-1)^2
> > as x -> 0 the original limit is
> > sum_{m=1}^infinity 1/(2m-1)^2 = pi^2/8.
>
> Very nice, thanks. One step I'm not sure about though - how do you
> show that the sum over m commutes with the limit x->0 in this case?
I was leaving that to you...but as an exercise show that the original
limit equals lim_{x->0+) of
sum_{k odd} (sinh x/2)^2/(sinh kx/2)^2
which is uniformly convergent since
(sinh y)/(sinh ky) <= 1/k.