Question

∑from 1to infinity of n^2* e^-2n

Answer 1

Does this converge or diverge?

Answer 2

converges like crazy

Answer 3

converges to?

Answer 4

\[\sum_{k=1}^\infty \frac{n^2}{e^{2n}}\]

Answer 5

who knows. just that the denominator grows tons faster than numerator. i will cheat and find the sum if you like

Answer 6

only if you can tell me how you got it? please

Answer 7

i have absolutely no idea how to add this up.

Answer 8

leave the question open and someone may know. i do know that the answer is
\[\frac{e^2(e^2+1)}{(e^2-1)^3}\]

Answer 9

alright thanks

Answer 10

but no idea how to sum. keep asking

Answer 11

this might help... http://www.wolframalpha.com/input/?i=%E2%88%91+from+1+to+infinity+of+n^2*+e^%28-2n%29

Answer 12

hey maya I think I know how to do this

Answer 13

give me a bit to work it out

Answer 14

ok I have the solution let me type it out:

Answer 15

great

Answer 16

let w=e^-2. Then our sum is:

\[\sum_{n=0}^{\infty}n^2w^n\]

now consider the following sum:
\[\sum_{n=0}^{\infty}w^n\]
we will take its second derivitive:
\[\left( \sum_{n=0}^{\infty} w^n\right) ''=\sum_{n=0}^{\infty}n*(n-1)*w^n=\sum_{n=0}^{\infty}n^2w^{n-2}-\sum_{n=0}^{\infty}nw^{n-2}\]

we know the geometric series' solution on the left so we can find its second derivative:
\[\sum_{n=0}^{\infty}w^n=\frac{1}{1-w}\]
\[\left( \frac{1}{1-w} \right)''=\frac{2}{(1-w)^3}\]
multiplying both sides by w^2

\[\frac{2w^2}{(1-w)^3}=\sum_{n=0}^{\infty}n^2w^n-\sum_{n=0}^{\infty}nw^n\]

Answer 17

ok almost there, now we have the sum you want in the equation

Answer 18

so we use the same trick for the rightmost sum:

\[\frac{w}{(1-w)^2}=\sum_{n=0}^{\infty}nw^n\]

now we plug in to find the sum we wanted:

\[\frac{2w^2}{(1-w)^3}+\frac{w}{(1-w)^2}=\sum_{n=0}^{\infty}n^2w^n\]

plug in e^-2=w and simplify and you get the answer satellite posted

Answer 19

ok so that is a lot to take in probably let me know if you have a question

Answer 20

comes out to:
\[\frac{e^2(1+e^2)}{(e^2-1)^3}\]

Answer 21

hmm trying this myself. I think i get it

Answer 22

thank you

Answer 23

you're welcome :)

Answer 24

prettyyy confusing would have never thought of doing that myself

Answer 25

we did stuff like this a lot in complex analysis that's the only reason I knew to do it.

Answer 26

oh small mistake when taking the second derivative of geometric series should be n*(n-1)*w^(n-2) not just w^n

Answer 27

just a typo

Answer 28

ok

Answer 29

WOW! what gave you the idea to take the second derivative??

Answer 30

trying to get the n^2 term in front :)

Answer 31

aha! very nice

Answer 32

thanks

Answer 33

actually very very nice

Answer 34

my complex analysis prof liked doing similar probs

Answer 35

really good prof, hes the one doing galois theory in the fall

Answer 36

then you will have lots of fun:)