Nov 29, 2011 #1 K KING Member May 16, 2008 191 0 16 ? (3x-2)^n / n3^n (up the sigma ? , and down the sigma n=0)
Nov 29, 2011 #2 G GilEspie New member Nov 29, 2011 1 0 1 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.
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.