how do i solve the sum from n=1 to infinity of (3n/1+8n)^n?

Question

do i use the root test, the limit test...which test?

Answer 1

Root test (Cauchy's):

(a_n)^(1/n) = ([3n/(1+8n)]^n)^(1/n)= 3n/(1+8n) = 3/(8+1/n)
----> 3/8 < 1, the series is convergent.

Answer 2

The sum of the mathematics series is given by utilising S =(n/2) (2a +(n-a million) d) the area a is the 1st term, n the form of words and d the version. in this subject a=2 d=2 n= ninety 5 So S = (ninety 5/2) ( 4 + ninety 4*2) S = (ninety 5/2) ( 192) S = ninety 5 * 86 = 8170

Answer 3

(3n/(1+8n))^n <-- I think you meant this. If so, you must show that the nth term is greater than the (n+1)th term and show
(3n/(1+8n))^n = 0
lim n--> infinity
If both of these criteria are met, then the series converges and has a sum.

Answer 4

thats a hard one to slove and im very good at math but if i knew i would give u the right answer right away

Answer 5

how can infinity be in an equation??????

Answer 6

no idea