? (3x-2)^n / n3^n (up the sigma ? , and down the sigma n=0)?

G

GilEspie

New member
I'm guessing that you want to test for convergence.

oo
? (3x-2)^n / n3^n
n=1

Let's use the Ratio Test.

lim n-->oo {(3x - 2)^(n+1) / [(n+1)3^(n+1)]} * [(n3^n) / (3x - 2)^n] =
lim n-->oo (3x - 2)/3 (n + 1) / n =
(3x - 2)/3 ilm n-->oo (1 + 1/n)/1 = 1

| 3x - 2 | < 1
-1 < 3x - 2 < 1
1 < 3x < 3
1/3 < x < 1 <== Radius of convergence.

Test your endpoints to check if the series converges at them.


Hope this helped.
 
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