Question

Help summing this series?

Answer 1

\[\sum_{1}^{\infty} n ^{2}(\frac{1}{3})^{n+1}\]

Answer 2

Wolfram says it's 4.5

http://www.wolframalpha.com/input/?i=Sum+1+to+infinity+n^2+%281%2F3%29^%28n-1%29

Answer 3

wolfram told me 1/2

Answer 4

n+1 not minus haha

Answer 5

http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity%28n^2%29%28%281%2F3%29^%28n%2B1%29%29

Answer 6

i was looking for how to do it by hand....

Answer 7

Sorry, can't see straight, 1/2 it is...

Answer 8

but how do i get to 1/2 ???

Answer 9

I have an inelegant solution, which this margin is too narrow to contain :p
Writing it out in a Word Doc instead, give me 5 minutes.

Answer 10

Never tried this before, hope it works:

Answer 11

inelegant... im a math dude, i have no idea what that means :)

Answer 12

It basically means it ain't pretty! On reflection, though, I don't think there's a neater way to do it - some maths problems just have inelegant solutions...

Answer 13

no no its actually very elegant hahaha
i was trying to do a similar thing but i kept it algebraic so i did not see the ability to arrange then classify as a geometric!
thanks so much!

Answer 14

another way:

\[\sum_{n=1}^{\infty}n^2\left(\frac{1}{3}\right)^{n+1}=\frac{1}{3^2}\sum_{n=1}^{\infty}n^2\left(\frac{1}{3}\right)^{n-1}\]

and consider
\[=\frac{1}{3^2}\sum_{n=1}^{\infty}n^2\left(x\right)^{n-1}\]

\[=\frac{1}{3^2}\sum_{n=1}^{\infty}n\cdot n\left(x\right)^{n-1}\]

\[=\frac{1}{3^2}\sum_{n=1}^{\infty}n\cdot (n+1-1)\left(x\right)^{n-1}\]

\[=\frac{1}{3^2}\sum_{n=1}^{\infty}(n\cdot (n+1)-n)\left(x\right)^{n-1}\]

\[=\frac{1}{3^2}\sum_{n=1}^{\infty}n\cdot (n+1)\left(x\right)^{n-1}-\frac{1}{3^2}\sum_{n=1}^{\infty}n\left(x\right)^{n-1}\]

integrate the first term twice and integrate the 2nd term once..find the sums...differentiate the appropriate number of times ...evaluate ate 1/3