Question

can anyone provide any guidance on this series?
Sum from n=0 to infinity of n((6^n)/n!)

Answer 1

i calculated the first few terms and got 0+6+108+216+324+... and i recognize that if that first n was not there then it would be e^6 but im not sure what to do with the n

Answer 2

haha hey again

Answer 3

howdy :)

Answer 4

e^6 is definantly part of it; do you know the answer already?

Answer 5

lol well i know worlfram alpha says it should be 6e^6

Answer 6

good, then the question is how to get there :)

Answer 7

correct i thought maybe you can do the sums seperatly then mulitply together but that did not work out

Answer 8

joes got a brilliant method im sure :)

Answer 9

Do you know the infinite series for e^x?

Answer 10

then i thought well i can end up getting n!/(n-1)! but that does not make sense for n=0

Answer 11

yes i do its x^n/n!

Answer 12

Take the infinite series for e^x:\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!}\]Take the derivative of each side:\[\frac{d}{dx}(e^x)=\frac{d}{dx}\left(\sum_{n=0}^{\infty }\frac{x^n}{n!}\right)=\sum_{n=1}^{\infty }\frac{nx^{n-1}}{n!}\]

Answer 13

This sorta matches what you want, but one thing is off.

Answer 14

Right now we are at:\[e^x=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!}\]The only thing that is off is the power. In your expression, you have a multiple of n, an n!, and an nth power. In mine, we have a multiple of n, an n!, and an (n-1)th power, so we need to fix that.

Answer 15

by multiplying it by x right?

Answer 16

thats correct :)

Answer 17

so we end up with:\[xe^x=\sum_{n=1}^\infty \frac{nx^n}{n!}\]

Answer 18

alright awesome so does it not matter that n goes from 1 to infinity as opposed to 0 to infinity?

Answer 19

like how did you know to do the derivative and if you did that then to change the limits?

Answer 20

at n=0 the add on is just zero right?

Answer 21

if you plug in n=0, regardless of the value of x, we will end up with 0, so there is no difference between n=0 to infinity or n=1 to infinity.

Answer 22

the derivative of e^x is equal to e^x

Answer 23

I knew the change the limits in the summation because the first term was a constant (the constant 1), and the derivative would kill it.

Answer 24

and the reason i choose to take a derivative in the first place is because your expression had an multiple of n in it, and derivatives bring down powers.

Answer 25

alright cool thank you very much i get it now

Answer 26

i really appreciate it