hello

i need to compute this sum:

sigma(n=1 to infinity) n^2 / 2^n

and i'm having trouble doing that.

i tried looking at a power series,but couldn't develop it.

help would be appreciated.

thanks.

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- August 7th 2006, 07:33 AMparallelcompute a sum
hello

i need to compute this sum:

sigma(n=1 to infinity) n^2 / 2^n

and i'm having trouble doing that.

i tried looking at a power series,but couldn't develop it.

help would be appreciated.

thanks. - August 7th 2006, 08:03 AMCaptainBlackQuote:

Originally Posted by**parallel**

a lot of the work yourself. If you have trouble we will fill in the detail.

Consider the geometric series:

which you should be able to sum for . Now differentiate

once and twice and see what you find after a bit of rearrangement of the

corresponding sums of the series, and letting .

RonL - August 7th 2006, 08:52 AMSoroban
Hello, parallel!

I just "invented" an approach to this problem . . .

Quote:

We have: .**[1]**

Multiply by :

. . . . . . .**[2]**

Subtract**[2]**from**[1]**:

. . . . . . .**[3]**

Multiply by :

. . . . . . .**[4]**

Subtract**[4]**from**[3]**:

. . . . . . .

. . . . . . .

. . . . . . . . . . . . . . an infinite geometric series with

. . . . . . . . . . . . . . . . . . Its sum is: .

Hence, we have: . . . . . (drumroll, please)

Therefore: . . . . . ta-*DAA!*

- August 7th 2006, 09:00 AMCaptainBlackQuote:

Originally Posted by**Soroban**

This is implicit in the method I outlined :D

CB - August 7th 2006, 09:30 AMparallel
thanks for the reply CaptainBlack,this is what I came up with,hope it's correct

the sum of (x/2)^n is : x/(2-x)

taking deriv:

(1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2

taking 2nd deriv:

sigma n(n-1)/2^n * x^(n-2)

this sum is:

(2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4

now I noticed that the second term is (1) * (1/x),

so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)

is this correct?

thanks for the help

Soroban,

hey, I really liked your solution,I'll sure keep this approach in mind :)

thanks for your help too. - August 7th 2006, 09:45 AMCaptainBlackQuote:

Originally Posted by**parallel**

with Soroban's so we may presume the algebra is correct.

This method is powerful as it provides a method of finding:

for any polynomial in (at least for

smallish polynomials) and .

RonL - August 7th 2006, 10:32 AMparallel
thanks again :)