Quiz 5 Quiz 5                              Friday, May 3, 2001



NAME:

Student Number:------            No.----           Marks ----






(25 Marks)     Find the convergence set for




n = 0 
(x-1)n
(n+1)2n

Solution. Let an = [1/((n+1)2n)] and an+1 = [1/((n+2)2n+1)]. Then [(an)/(an+1)] = [(2(n+1))/(n+1)] and


r =
lim
n 
an
an+1
=
lim
n 
2(n+1)
n+1
= 2.

Hence, we have

(i) when |x-1| < 2 (or -1 < x < 3), the series converges absolutely.

(ii) When |x-1| > 2 (or x < -1 or x > 3), the series diverges.

(iii) When x-1 = 2 (or x = 3), the series becomes n = 0[1/(n+1)] and diverges.

(iv) When x-1 = -2 (or x = -1), the series becomes n = 0[((-1)n)/(n+1)]. It follows from the Alternating Series Test that n = 0[((-1)n)/(n+1)] converges.

Hence, the convergence set is -1 x < 3 or [-1, 3).


File translated from TEX by TTH, version 2.79.
On 12 May 2001, 09:19.