Question

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
pi^2/(3^n+1)

Answer 1

Having trouble with these concepts for some reason..

Answer 2

Im sorry its supposed to be pi^n

Answer 3

pi^n/(3^(n+1)) = (1/3) * (pi/3)^n correct?

Answer 4

im not sure how you got that..

Answer 5

(1/3)(1/3^n) = 1/(3^(n+1))

Answer 6

a^b*a^c = a^(b+c)

Answer 7

ok yes.

Answer 8

what is the common ratio for
(1/3) * (pi/3)^n
?

Answer 9

a(b)^n then b is common ratio

Answer 10

pi/3^n? and a =1/3?

Answer 11

ratio is pi/3

Answer 12

pi/3 > 1 so it does not converge

Answer 13

oh ok right.

Answer 14

this was all about the algebra to get it into the form a(b)^n
get used to factoring things in/out of the denominator to increase/decrease the exponent

Answer 15

its a handy "trick"

Answer 16

oh ok...im sorry...how did you get pi/3^n again?

Answer 17

is that n+1 an exponent \(k^{(n+1)}\)? or is it \(k^{(n)}+1\)

Answer 18

its the first one.

Answer 19

then zzr has this taken care of :)

Answer 20

(pi)^n/3^(n+1) = (pi)^n * (1/3^(n+1)) = (pi)^n * (1/3) * (1/3^n) = (1/3) * (pi)^n/3^n = (1/3) (pi/3)^n

Answer 21

oh ok I got it! I was just having alittle trouble deciphering what you were saying...I get it now!

Answer 22

\[\frac{\pi^n}{3^{n+1}}= (1/3)\frac{\pi^n}{3^n}= (1/3)(\frac{\pi}{3})^n\]

Answer 23

i get it! Thanks alot zzr

Answer 24

ok sorry, the equation editor was not working for me but is now:)

Answer 25

np