Question

find the exact sum of the series

Answer 1

where is the series?

Answer 2

As there is no series, so sum is zero..

Answer 3

no sum is imaginary

Answer 4

When there is not any series how you can say that it is imaginary??

Answer 5

\[\sum_{n=1}^{\infty} x^2e^(-x)\]

Answer 6

when there is no seires how can you say it is zero?

Answer 7

Values of Empty Set is 0..

Answer 8

there are no 'n''s in your sum

Answer 9

n=1

Answer 10

x^2e^(-x)

Answer 11

I think you need to reexamine your problem to make sure you typed it correctly

Answer 12

\[\sum_{n=1}^{\infty} n^3e^(-n)\]

Answer 13

sorry

Answer 14

dont i first compare it to a similar equation say f(x)=x^3e^(-x) then it is continuous and decreasing and f(x)>0

Answer 15

that won't help you find the exact sum

Answer 16

okay so what do i need to do?

Answer 17

I don't see a nice way to do this...

I would start by writing it as

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

Answer 18

have you done anything with differentiating and integrating a series?

Answer 19

yes

Answer 20

good ...then start integrating ;)

Answer 21

\[n^4/4? \] then isnt 1/e the lnx?

Answer 22

you need to get rid of the \(n^3\)...you will have to do it more than once

Answer 23

\[n^3x^n\]

\[x\cdot n^3x^{n-1}\]

integrate

\[n^3x^{n-1}\]

Answer 24

n^2x^n

Answer 25

yes

Answer 26

now you need to keep woking it so the \(n^2\) is gone

Answer 27

\[n^2x^(n+1)/n+1\]

Answer 28

that doesn't help you reduce the \(n^2\)

Answer 29

I've figured out a slicker way of doing it..though similar

..

Answer 30

notice that n\[n^3=n^3-n+n\]

and \[n^3-n=n(n^2-1)=n(n+1)(n-1)=(n+1)n(n-1)\]

so \[\sum_{n=1}^{\infty}n^3x^n=\sum_{n=1}^{\infty}(n+1)n(n-1)x^n+\sum_{n=1}^{\infty}nx^n\]

Answer 31

now use the technique you started using above...

this is not a pretty problem ;)

Answer 32

ok i see what you did you factored out

Answer 33

after all is said and done...I get

\[\frac{e \left(1+4 e+e^2\right)}{(e-1)^4}\] as the final answer...but it is a nasty process

Answer 34

because after you integrate a few times...you will have to differentiate.

Answer 35

ok but that doesnt look much like the solutions to the problems we have done in class

Answer 36

i mean we are supposed to use the integral test to show that it converges

Answer 37

if that is all you want to test is convergence then the integral test will do...if you want the exact value of the sum you will have to do what i showed above

Answer 38

you asked for the "exact sum of the series"

Answer 39

true let me open the correct question im sorry