Transcription of Math 104: Introduction to Analysis SOLUTIONS
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Math 104: Introduction to Analysis SOLUTIONS
Math 104: Introduction to AnalysisSOLUTIONSA lexander GiventalHOMEWORK that12+ 22+ +n2=16n(n+ 1)(2n+ 1)for alln (n) =n(n+ 1)(2n+ 1)/6. Thenf(1) = 1, the theoremholds true forn= 1. To prove the theorem, it suffices to assume thatit holds true forn=mand derive it forn=m+ 1,m= 1,2,3, ..We havef(m+ 1) f(m) =16(m+ 1)[(2m+ 3)(m+ 2) m(2m+ 1)]=16(m+ 1)(6m+ 6) = (m+ 1) the induction hypothesis,f(m) = mk=1k2, and thereforef(m+ 1) =f(m) + (m+ 1)2=m+1 k= for whichnthe inequality2n> n2holds true, and proveit by mathematical inequality is falsen= 2,3,4, and holds true for all othern , it is true by inspection forn= 1, and the equality 24= 42holds true forn= 4. Thus, to prove the inequality for alln 5, itsuffices to prove the following inductive step:For anyn 4, if 2n n2, then 2n+1>(n+ 1) is not hard to see: 2n+1= 2 2n 2n2, which is greater than(n+1)2provided that (n+1)< whenn >1/( 2 1) = 2+1,which includes all integersn (nk):=n!
4 Applying other theorems about behavior of limits under arithmetic operations with sequences, we conclude that lim 1 2 q 1+ 1 4n +2 = 1 2·1+2 = 1 4. 9.5. Let t1 = 1 and tn+1 = (t2 n + 2)/2tn for n ≥ 1. Assume that tn converges and find the limit.
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