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 doneOriginally 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
Hello, parallel!
I just "invented" an approach to this problem . . .
$\displaystyle \sum^{\infty}_{n=1}\frac{n^2}{2^n}$
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!
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.
That's what I had in mind. I will not check it in detail as the answer agreesOriginally 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