## anonymous one year ago 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.

1. Jhannybean

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

2. anonymous

haha yes

3. jdoe0001

hmmm dunno that one :(

4. amistre64

does the sequence limit to zero?

5. amistre64

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

6. anonymous

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

7. anonymous

oh sorry it's over x!

8. amistre64

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

9. anonymous

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

10. amistre64

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

11. anonymous

will try, h/o

12. anonymous

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

13. amistre64

pdf, probability density function ...

14. anonymous

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

15. anonymous

$x = 0,1,2,...$

16. amistre64

we want the sumto equal ...1?

17. anonymous

yes

18. anonymous

i know i can pull out the constant

19. amistre64

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

20. amistre64

sum of ...

21. amistre64

do you recall the taylor series for e^x ?

22. anonymous

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

23. amistre64

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

24. anonymous

ah, yes. that looks like it exactly

25. anonymous

$e^x = \sum \frac{x^n}{n!}$

26. amistre64

im so clever it scares me ....

27. anonymous

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

28. anonymous

i appreciate the help

29. amistre64

youre welcome :)