HW 1 Solutions

HW 1 Solutions - To prove the converse, assume now only...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 117, F 09, HW#1 (sample solutions) P.54, #3. Let y ( x - ±, x + ± ). Then x - ± < y < x + ± , and so the differences y - ( x - ± ) and ( x + ± ) - y are both positive. Set δ = min { y - ( x - ± ) , ( x + ± ) - y } , and note that δ > 0. On one hand, δ y - ( x - ± ) = y - x + ± , and so x - ± y - δ . On the other hand, δ x + ± - y , and so y + δ x + ± . To summarize: x - ± y - δ < y < y + δ x + ±. Consequently, ( y - δ, y + δ ) ( x - ±, x + ± ). This proves that ( x - ±, x + ± ) is a neighborhood of y . ± P.55, #6d. Claim: The sequence { 3 n 2 n +1 } n =1 converges to 3 2 . Given a real number ± > 0, choose a positive integer N > 1 4 ( 3 ± - 2). For any integer n N , we then have n > 1 4 ( 3 ± - 2), which implies 4 n > 3 ± - 2, which in turn implies 4 n + 2 > 3 ± . This implies (4 n + 2) ± > 3 (because ± > 0), whence ± > 3 4 n +2 (because 4 n + 2 > 0). Thus, ± ± ± ± 3 n 2 n + 1 - 3 2 ± ± ± ± = ± ± ± ± - 3 4 n + 2 ± ± ± ± = 3 4 n + 2 < ±. Since this holds for all n N , we have proved that the given sequence does converge to 3 2 . ± P.55, #7. Assume first that { a n } n =1 converges to A . Then given any ± > 0, there exists a positive integer N such that | a n - A | < ± for all n N . We can rewrite this information as | ( a n - A ) - 0 | < ± for all n N . This proves that the sequence { a n - A } n =1 converges to 0.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: To prove the converse, assume now only that { a n-A } n =1 converges to 0. Then given any &gt; 0, there exists a positive integer N such that | ( a n-A )-| &lt; for all n N . This just says that | a n-A | &lt; for all n N . Therefore { a n } n =1 converges to A . P.55, #9. Let &gt; 0. Since { a n } n =1 converges to A , there exists a positive integer N a such that | a n-A | &lt; for all n N a . Moreover, since { b n } n =1 converges to A , there exists a positive integer N b such that | b n-A | &lt; for all n N b . Set N = max { N a , N b } . Then we have both | a n-A | &lt; and | b n-A | &lt; for all n N . Since a n c n b n for all n , we thus have A- &lt; a n c n b n &lt; A + for all n N , and so | c n-A | &lt; for all n N . Therefore the sequence { c n } n =1 converges to A ....
View Full Document

Ask a homework question - tutors are online