Rudin_3-1

# Rudin_3-1 - Numerical Sequences and Series Written by Men-Gen Tsai email [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ 1 Prove that the convergence of{sn implies

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Numerical Sequences and Series Written by Men-Gen Tsai email: [email protected] 1. Prove that the convergence of { s n } implies convergence of {| s n |} . Is the converse true? Solution: Since { s n } is convergent, for any ± > 0, there exists N such that | s n - s | < ± whenever n N . By Exercise 1.13 I know that || s n | - | s || ≤ | s n - s | . Thus, || s n | - | s || , that is, { s n } is convergent. The converse is not true. Consider s n =( - 1) n . 2. Calculate lim n →∞ ( n 2 + n - n ). Solution: n 2 + n - n = n n 2 + n + n = 1 ± 1 /n + 1 + 1 1 2 as n →∞ . 3. If s n = 2 and s n +1 = ± 2+ s n ( n =1 , 2 , 3 ,... ) , prove that { s n } converges, and that s n < 2 for n , 2 , 3 . Proof: First, I show that { s n } is strictly increasing. It is trivial that s 2 = ± s 1 = ² ± 2 > 2= s 1 . Suppose s k >s k - 1 when 1

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k < n . By the induction hypothesis, s n = ± 2+ s n - 1 > ± s n - 2 = s n - 1 By the induction, { s n } is strictly increasing. Next, I show that { s n } is bounded by 2. Similarly, I apply the induction again. Hence { s n } is strictly increasing and bounded, that is, { s n } converges. 4. 5. 6. 7. Prove that the convergence of a n implies the convergence of ² a n n if a n 0. Proof: By Cauchy’s inequality, k ² n =1 a n k ² n =1 1 n 2 k ² n =1 a n a n n for all n N . Also, both a n and 1 n 2 are convergent; thus k n =1 a n a n n is bounded. Besides, a n n 0 for all n . Hence a n n is convergent. 8. 9. Find the radius of convergence of each of the following power series: ( a ) ² n 3 z n , ( b ) ² 2 n n ! z n , ( c ) ² 2 n n 2 z n , ( d ) ² n 3 3 n z n . 2
Solution: (a) α n =( n 3 ) 1 /n 1 as n →∞ . Hence R =1 = 1. (b) α n = (2 n /n !) 1 /n =2 / ( n !) 1 /n 0 as n . Hence R =+ . (c) α n = (2 n /n 2 ) 1 /n 2 / 1 = 2 as n . Hence R / 2. (d) α n n 3 / 3 n ) 1 /n 1 / 3 as n . Hence R = 3. 10. 11. Suppose a n > 0, s n = a 1 + ... + a n , and a n diverges. (a) Prove that a n 1+ a n diverges. (b) Prove that a N +1 s N +1 + ... + a N + k s N + k 1 - s N s N + k and deduce that a n s n diverges. (c) Prove that a n s 2 n 1 s n - 1 - 1 s n and deduce that a n s 2 n converges. (d) What can be said about ± a n 1+ na n and ± a n n 2 a n ? Proof of (a): Note that a n a n 0 1 1 a n +1 0 1 a n a n 0 as n . If a n 1+ a n converges, then a n 0 as n . Thus for some ± ± = 1 there is an N 1 such that a n < 1 whenever n N 1 . Since a n 1+ a n converges, for any ± > 0 there is an N 2 such that a m a m + ... + a n a n < ± 3

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all n>m N 2 . Take N = max( N 1 ,N 2 ). Thus ± > a m 1+ a m + ... + a n a n > a m 1+1 + ... + a n = a m + ... + a n 2 for all N . Thus a m + ... + a n < 2 ± for all n > m N . It is a contradiction. Hence a n 1+ a n diverges. Proof of (b): a N +1 s N +1 + ... + a N + k s N + k a N +1 s N + k + ... + a N + k s N + k = a N +1 + ... + a N + k s N + k = s N + k - s N s N + k =1 - s N s N + k If a n s n converges, for any ± > 0 there exists N such that a m s m + ... + a n s n < ± for all m,n whenever n > m N . Fix m = N and let n = N + k . Thus ± > a m s m + ... + a n s n = a N s N + ... + a N + k s N + k 1 - s N s N + k 4
for all k N . But s N + k →∞ as k since a n diverges and a n > 0. Take ± =1 / 2 and we obtain a contradiction. Hence a n s n diverges.

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## This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.

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Rudin_3-1 - Numerical Sequences and Series Written by Men-Gen Tsai email [email protected] 1 Prove that the convergence of{sn implies

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