Midterm Exam 1 Practice Questions with Solutions - Analysis I | MATH 313, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Midterm Exam 1 Practice Questions with Solutions - Analysis I | MATH 313 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Solutions – Practice Questions Midterm 1 – Math 313 – Fall 2010 1. We have for every n ∈ N an = 3 + 13n 1− 2n2 . (1) Set bn := 3 and cn := 1n . It follows that bn → 3 and cn → 0. Therefore, by the product theorem, 13cn → 0 and, by the sum theorem, 3 + 13n = bn + 13cn → 3. Similarly, 1− 2 n2 → 1. Therefore, by (1) and the quotient theorem, an → 3. 2. For each n ∈ N we have an = n! nn = 1 n · 2 n · · · · · n− 1 n · n n . Since 0 < in ≤ 1 for every i = 1, . . . , n we obtain 0 < an ≤ 1 n for every n ∈ N. It now follows easily that an converges to 0. Indeed, let ε > 0. There exists n0 ∈ N such that 1n0 ≤ ε. Then |an| = an ≤ 1 n ≤ 1 n0 ≤ ε for every n ≥ n0. 3. We first note that for every n ∈ N we have an = n− √ n2 + 4n + 3 = (n− √ n2 + 4n + 3) · (n + √ n2 + 4n + 3) n + √ n2 + 4n + 3 = n2 − (n2 + 4n + 3) n(1 + √ 1 + 4n + 3 n2 ) = −4n− 3 n(1 + √ 1 + 4n + 3 n2 ) = −4− 3n 1 + √ 1 + 4n + 3 n2 . It is not difficult to see that if a sequence (bn) with bn ≥ 0 converges to some b > 0 then √ bn converges to √ b. Therefore, √ 1 + 4n + 3 n2 converges to 1. Since −4 − 3 n converges to −4 it follows from the theorem about quotients of sequences that an converges to −42 = −2. Remark: We prove the above claim that if a sequence (bn) with bn ≥ 0 converges to some b > 0 then √ bn converges to √ b. Indeed, | √ b− √ bn| = ∣∣∣∣∣ ( √ b− √ bn)( √ b + √ bn)√ b + √ bn ∣∣∣∣∣ = |b− bn|√b +√bn ≤ 1√b |b− bn|. Now, given ε > 0 there exists n0 ∈ N such that |b− bn| ≤ ε √ b for all n ≥ n0. It then follows that | √ b− √ bn| ≤ ε for all n ≥ n0. This proves the claim.