Math 5333 Solns-Assignment4 - Solutions Homework#4 Section 4.2 6(a Counterexample(sn =(1 1 1 1(tn =(1 1 1 1(b Counterexample the example in part(a will

Math 5333 Solns-Assignment4 - Solutions Homework#4 Section...

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Solutions, Homework #4 Section 4.2 6. (a) Counterexample: ( s n ) = (1 , - 1 , 1 , - 1 , · · · ); ( t n ) = ( - 1 , 1 , - 1 , 1 , · · · ). (b) Counterexample: the example in part (a) will work here, too. (c) Proof: Suppose ( s n ) A and ( s n + t n ) B . Then t n = ( s n + t n ) - s n B - A (d) Counterexample: ( s n ) = parenleftbigg 1 , 1 2 , 1 3 , · · · , 1 n · · · parenrightbigg , s n 0, ( t n ) = (( - 1) n ) = ( - 1 , 1 , - 1 , 1 , . . . ) does not converge. But ( s n t n ) = parenleftbigg ( - 1) n n parenrightbigg 0. 8. (a) There are lots of examples. Here are two: ( s n ) = parenleftbigg 1 n parenrightbigg , ( s n ) = parenleftbigg n n 2 + 1 parenrightbigg (b) There are lots of examples. Here are two: ( s n ) = ( n ) = (1 , 2 , 3 , 4 , . . . ) ( s n ) = parenleftbigg n 2 n + 1 parenrightbigg 9. ( s n ) and ( t n ) are sequences such that s n t n for all n . (a) Choose any number M R . Since lim n →∞ s n = + , there is a positive integer N such that s n > M for all n > N . Since t n s n for all n , it follows that t n > M for all n > N . Therefore lim n →∞ t n = + : (b) Choose any number M R . Since lim n →∞ t n = -∞ , there is a positive integer N such that t n < M for all n > N . Since s n t n for all n , it follows that s n < M for all n > N . Therefore lim n →∞ s n = -∞ : 12. Let epsilon1 > 0. Since ( s n ) is convergent, it is bounded; | s n | ≤ M for all n . 1
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| s 2 n - s 2 | = | s n - s | | s n + s | ≤ | s n - s | ( M + | s | ). Since s n s there is a positive integer N such that | s n - s | < epsilon1 M + | s | for all n > N . Therefore vextendsingle vextendsingle s 2 n - s 2 vextendsingle vextendsingle = | s n - s | | s n + s | ≤ | s n - s | ( M + | s | ) < ( M + | s | ) epsilon1 M + | s | = epsilon1 for all n > N and it follows that s 2 n s 2 . 15. (a) n + 1 - n = parenleftBig n + 1 - n parenrightBig n + 1 + n n + 1 + n = 1 n + 1 + n < 1 2 n 0 (b) n 2 + 1 - n = parenleftBig n 2 + 1 - n parenrightBig n 2 + 1 + n n 2 + 1 + n = 1 n 2 + 1 + n < 1 2 n 0 (c) n 2 + n - n = parenleftBig n 2 + n - n parenrightBig n 2 + n + n n 2 + n + n = n n 2 + n + n 1 2 17. Let s n = k n /n !.
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