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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- Aug 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. - Aug 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:

$\displaystyle

S(x)=\sum_1^{\infty} (x/2)^n

$

which you should be able to sum for $\displaystyle |x|<2$. Now differentiate

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

corresponding sums of the series, and letting $\displaystyle x=1$.

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

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

Quote:

$\displaystyle \sum^{\infty}_{n=1}\frac{n^2}{2^n}$

We have: .$\displaystyle S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots$**[1]**

Multiply by $\displaystyle \frac{1}{2}$:

. . . . . . . $\displaystyle \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots$**[2]**

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

. . . . . . . $\displaystyle \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots$**[3]**

Multiply by $\displaystyle \frac{1}{2}$:

. . . . . . . $\displaystyle \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots$**[4]**

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

. . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots$

. . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$

. . . . . . . . . . . . . . an infinite geometric series with $\displaystyle a = \frac{1}{2},\;r = \frac{1}{2}$

. . . . . . . . . . . . . . . . . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$

Hence, we have: .$\displaystyle \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2}$ . . . . (drumroll, please)

Therefore: .$\displaystyle \boxed{S\;=\;6}$ . . . . ta-*DAA!*

- Aug 7th 2006, 09:00 AMCaptainBlackQuote:

Originally Posted by**Soroban**

This is implicit in the method I outlined :D

CB - Aug 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. - Aug 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:

$\displaystyle

S=\sum_1^{\infty}P(n)/k^n

$

for any polynomial $\displaystyle P(n)$ in $\displaystyle n$ (at least for

smallish polynomials) and $\displaystyle k>1$.

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