Question

I'm trying to find what the summation from 0 to infinity of ((e^-4)4^x)/4! converges to (trying to show it is a valid pdf). Any suggestions on how to approach this would be welcome.

Answer 1

\[\huge \sum_{0}^{\infty} \frac{4^x(e^{-4})}{4!}\] Is it this? haha

Answer 2

haha yes

Answer 3

hmmm dunno that one :(

Answer 4

does the sequence limit to zero?

Answer 5

e^(-4)/4! is just some constant value ...

Answer 6

I believe it does, since the factorial grows faster than the exponential

Answer 7

oh sorry it's over x!

Answer 8

4^x is the only thing that varies .. the constant factors out ...

Answer 9

you're right, without that it wouldn't converge

Answer 10

.... can you pic the question? avoids alot of error

Answer 11

will try, h/o

Answer 12

\[p(x) = \frac{e^{-4}4^x}{x!}\]

Answer 13

pdf, probability density function ...

Answer 14

yeah, pmf not pdf. sorry, the pic was coming out really grainy

Answer 15

\[x = 0,1,2,...\]

Answer 16

we want the sumto equal ...1?

Answer 17

yes

Answer 18

i know i can pull out the constant

Answer 19

then it behooves us to consider that 4^x/x! = e^(4)

Answer 20

sum of ...

Answer 21

do you recall the taylor series for e^x ?

Answer 22

i don't i can look it up, that may be the prod in the right direction i am looking for

Answer 23

its a thought, not sure how well it will turn out tho

Answer 24

ah, yes. that looks like it exactly

Answer 25

\[e^x = \sum \frac{x^n}{n!}\]

Answer 26

im so clever it scares me ....

Answer 27

haha i liked how you looked at it in terms of solving for what would give us e^4

Answer 28

i appreciate the help

Answer 29

youre welcome :)