Question

summation of 5^n/e^n-3n from 1 to infinity

Answer 1

\[\sum_{n=1}^{\infty}\frac{5^n}{e^n-3n}\]

Answer 2

is that the problem?

Answer 3

Yes
determine convergence

Answer 4

Do I use the geometric series test?

Answer 5

and i take it this is a power series that is comprised of the sequence created by the formula... as opposed to the sequence itself right?

Answer 6

Yess

Answer 7

maybe an integral test could be applied ...

Answer 8

could i make r=(5/e)^n?

Answer 9

i dont see how you could; what are the steps you would take to determine that?

Answer 10

i dk i ui guessed cause e is a constant. I'll try integral

Answer 11

But it's not really a good function to integrate. what would i use to integrate that? there's so many constants

Answer 12

divide it all by e^n is what i think you tried there

\[\sum_{n=1}^{\infty}\frac{5^n/e^n}{e^n/e^n-3n/e^n}\]

\[\sum_{n=1}^{\infty}\frac{\cfrac{5^n}{e^n}}{1-\cfrac{3n}{e^n}}\]

\[\sum_{n=1}^{\infty}\frac{\left(\cfrac{5}{e}\right)^n}{1-\cfrac{3n}{e^n}}\]

Answer 13

this would probably conform to \(\cfrac{a}{1-r}\), but I dont have to much experience at it to be sure

Answer 14

the top is a constant that does to infinity ... so my gut says it diverges, but i cant recall it correctly to be sure

Answer 15

what if i multiply by the reciproical and cancel the e^n's.....

Answer 16

then i have r=5/3 which would diverge....

Answer 17

but my gut says that's wrong

Answer 18

3n/e^n at n=inf goes to zero leaving us with:
\[\frac{5}{e}\ ^n\text{ ; at n=inf}\]
so the sequence doesnt converge, therefore the series doesnt converge .. id say divergant on all accounts

Answer 19

http://www.wolframalpha.com/input/?i=sum%285^n%2Fe^n-3n%29+from+1+to+infinity

Answer 20

or even with the correct equation :)

http://www.wolframalpha.com/input/?i=sum%285^n%2F%28e^n-3n%29%29+from+1+to+infinity

Answer 21

Thanks but to prove it i need to identify bn and take the limit of both an and bn. what do I set as bn?

Answer 22

the only thing i recall on that is to do partial decomp; and i dont quite see how to accomplish that on this one

Answer 23

bn=5^n/e^n. n-n=1....

Answer 24

i think ive used up all me wits that i can afford on this one :) good luck with it