analysis practice class test 2014

2
 A n al ys i s P r ac t i ce C l as s T es t 20 14 / 15 : 1 . I f ( a n ) is a se q u en ce o f re al nu m b er s; d e n e w h at i s m e an t by sa yi ng t h a t ( a n ) co n ve r g e s to t h e l imit l  as n  g oes to i n n it y. G iven ε  >0, ther e e xi st s N ϵ   N  su ch t h a t | a n  - l  | < ε  f o r a ll n >N . P r o ve d ir e ctly fr o m t h e d e n i t ion o f co n ve rge n ceth a t n 3 3 2 n 3 30 1 2  as n  Let ε  > 0 b e given. C hoo se N ϵ   N  such that N> m ax { 3, 3 12 3  } T hen f or n > N w e have: |  n 3 3 2 n 3 30 1 2 |  = | n 3 3n 3 +15 2n 3 30  | = 12 | 2 n 3 30 |  < 12 n 3  < 12  N 3 <ε  (f o r n 4 )   1 1 1 2  3

Upload: ac

Post on 07-Oct-2015

213 views

Category:

Documents


0 download

DESCRIPTION

Class test on Analysis

TRANSCRIPT

Analysis Practice Class Test 2014/15:

1. If () is a sequence of real numbers; define what is meant by saying that () converges to the limit as goes to infinity. 1 1 1

Given > 0, there exists N such that | - | < for all n > N.

Prove directly from the definition of convergence that as

Let > 0 be given. Choose N such that N > max 1 2

3

Then for n > N we have: = = < < (for n4)

2. Show that the following sequences converge to the given limits. You may use the theorems proved in lectures, but you must explain why any hypotheses they require hold.

(a) as

As for , we have by the AOL. 1 1 1

(b) as

As as and