{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10 - SOLUTIONS FOR HOMEWORK 5 10.1(a an = 1/n 1 = a1 > a2 >...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTIONS FOR HOMEWORK 5 10.1. (a) a n = 1 /n . 1 = a 1 > a 2 > . . . > 0, hence this sequence is non-increasing and bounded. (b) b n = ( 1) n /n 2 . This sequence is bounded, since 1 b n 1 for any n . However, it is not monotone: b 1 < b 2 , yet b 2 > b 3 . (c) c n = n 5 . This sequence is non-decreasing ( c 1 < c 2 < . . . ), and unbounded (by Example 9.7 and Theorem 9.10, lim c n = + ). (d) d n = sin( πn/ 7). This sequence is bounded: 1 d n 1 for every n . However, this sequence is not monotone: d n = d n +14 for every n , hence d 1 < d 2 , yet d 2 > d 15 . (e) e n = ( 2) n . The sequence is not monotone ( e 1 < e 2 , e 3 < e 2 ), and unbounded (lim | e n | = + ). (f) f n = n/ 3 n . We prove that this sequence is non-increasing. Indeed, f n +1 f n = 1 3 · n + 1 n = 1 3 parenleftBig 1 + 1 n parenrightBig 1 3 parenleftBig 1 + 1 parenrightBig < 1 . Thus, 1 / 3 = f 1 > f 2 > f 3 > . . . > 0, and our sequence is bounded. 10.6. (a) For every ε > 0 there exists N N s.t. 1 / 2 N < ε (see Example 9.7). We shall show that | s n s m | < ε whenever n, m > N . Indeed, suppose m > n > N . By the triangle inequality, | s m s n | = vextendsingle vextendsingle vextendsingle m - 1 summationdisplay k = n ( s k +1 s k ) vextendsingle vextendsingle vextendsingle m - 1 summationdisplay k = n | s k +1 s k | < m - 1 summationdisplay k = n 1 2 k = 1 / 2 n 1 / 2 m 1 1 / 2 < 1 2 n - 1 1 2 N < ε. (b) For n N let s n = n summationdisplay k =1 1 k .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}