# compute a sum

• Aug 7th 2006, 08:33 AM
parallel
compute 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, 09:03 AM
CaptainBlack
Quote:

Originally Posted by parallel
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.

This is a bit cryptic, but if you can do it from this you will have done
a lot of the work yourself. If you have trouble we will fill in the detail.

Consider the geometric series:

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

which you should be able to sum for $|x|<2$. Now differentiate
once and twice and see what you find after a bit of rearrangement of the
corresponding sums of the series, and letting $x=1$.

RonL
• Aug 7th 2006, 09:52 AM
Soroban
Hello, parallel!

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

Quote:

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

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

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

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

Subtract [2] from [1]:

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

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

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

Subtract [4] from [3]:

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

. . . . . . . $\frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
. . . . . . . . . . . . . . an infinite geometric series with $a = \frac{1}{2},\;r = \frac{1}{2}$
. . . . . . . . . . . . . . . . . . Its sum is: . $\frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$

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

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

• Aug 7th 2006, 10:00 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, parallel!

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

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

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

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

Subtract [2] from [1]:

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

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

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

Subtract [4] from [3]:

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

. . . . . . . $\frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
. . . . . . . . . . . . . . an infinite geometric series with $a = \frac{1}{2},\;r = \frac{1}{2}$
. . . . . . . . . . . . . . . . . . Its sum is: . $\frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$

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

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

This is implicit in the method I outlined :D

CB
• Aug 7th 2006, 10:30 AM
parallel
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 :)
• Aug 7th 2006, 10:45 AM
CaptainBlack
Quote:

Originally Posted by parallel
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 :)

That's what I had in mind. I will not check it in detail as the answer agrees
with Soroban's so we may presume the algebra is correct.

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

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

for any polynomial $P(n)$ in $n$ (at least for
smallish polynomials) and $k>1$.

RonL
• Aug 7th 2006, 11:32 AM
parallel
thanks again :)