Yllona Vanessy

Posted on Wednesday, 29 October, 2003 - 11:44 pm:

¥
å
n=1

(n²/2ⁿ)

It is supposed to be 6. I have no idea how to get to this convergent value.

David Loeffler

Posted on Thursday, 30 October, 2003 - 01:01 am:

This question is a lot easier if you use calculus. You can do without it, but it's rather harder work.

I presume you know that for any x,
¥
å
n=0
xⁿ = 1/(1-x)

. Try differentiating both sides of this equation: does this help at all?

David

Andre Rzym

Posted on Thursday, 30 October, 2003 - 10:24 am:

If you want to avoid calculus:

Define

S= ¥
å
n=1
n²/2ⁿ

Then

S-S/2= ¥
å
n=1
n²/2ⁿ - n²/2ⁿ⁺¹

S-S/2=1/2+ ¥
å
n=2
[n²-(n+1)²]/2ⁿ

S-S/2=1/2+ ¥
å
n=2
[2n-1]/2ⁿ

The latter sum splits into
2 ¥
å
n=2
n/2ⁿ

and
¥
å
n=2
1/2ⁿ

. You probably know how to do the latter sum, and the former sum can be written in terms of the latter by the same technique that was used on S.

Andre

Kerwin Hui

Posted on Thursday, 30 October, 2003 - 04:13 pm:

The cheapest way to get this value is to observe this is the expected value of X^2, where X is a geometric random variable with parameter p=1/2. Recall the mean and variance of geometric r.v. of parameter p is given by 1/p and q/p² respectively.

Kerwin

David Loeffler

Posted on Thursday, 30 October, 2003 - 04:37 pm:

And how do you propose to prove that without evaluating precisely this series, Kerwin?

Yatir Halevi

Posted on Thursday, 30 October, 2003 - 06:18 pm:

David, using your method I got a sum for:
n² /2^{2- n}
Maybe I'm a little rusty but how do you continue?

Yatir

BTW How are you all doing?

Andre Rzym

Posted on Thursday, 30 October, 2003 - 07:22 pm:

Yatir are you evaluating your formula at x=2 rather than for x=1/2 perchance?

Andre

Kerwin Hui

Posted on Thursday, 30 October, 2003 - 07:24 pm:

Consider x=kz (some fix k) and do the differentiation wrt z, not x. An appropriate choice of k gives the sum for n² /2ⁿ .

Kerwin

Yatir Halevi

Posted on Thursday, 30 October, 2003 - 10:30 pm:

hmmm...Stupid me.

Yatir

Andre Rzym

Posted on Thursday, 30 October, 2003 - 10:43 pm:

where the latter two expressions are evaluated at x=1/2

Andre

Yllona Vanessy

Posted on Thursday, 30 October, 2003 - 11:27 pm:

ok thanks guys. I tried David and Andre and i got it. Thanks again for your help.