AnalysisExam2 - Real Analysis Exam II Extra Credit David Joseph Stith 3 Problem Let(xn be the sequence dened recursively by x1 = 2 and xn 1 = for all n

# AnalysisExam2 - Real Analysis Exam II Extra Credit David...

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Real Analysis: Exam II Extra Credit David Joseph Stith 3. Problem. Let ( x n ) be the sequence defined recursively by x 1 = 2 and x n +1 = x n 2 + 5 x n for all n 1 . Show that ( x n ) converges and find the limit of the sequence. To show that ( x n ) converges, we will show that ( x n ) is bounded below and that ( x n ) is ultimately decreasing. It will then follow that ( x n ) is bounded above as well and hence, by the Monotone Convergence Theorem, converges. To show that ( x n ) is bounded below, we will show by Mathematical Induction that x n > 0 for all n N . We have that x 1 = 2 > 0. Suppose x k > 0 for some k N . Then x k / 2 > 0 and 5 /x k > 0 so that x k +1 = x k 2 + 5 x k > 0. Therefore x k > 0 = x k +1 > 0. Therefore by Mathematical Induction, x n > 0 for all n N . Therefore ( x n ) is bounded below. Furthermore we can easily see that x n R for all n N . Now we will show that x n +1 x n +2 for all n N and hence that the one-tail of ( x n ) is decreasing. Let n N . Then x n +1 = x n 2 + 5 x n = x n +1 = x 2 n + 10 2 x n = x 2 n - 2 x n x n +1 + 10 = 0 = x 2 n - 2 x n x n +1 = - 10 = x 2 n - 2 x n x n +1 + x 2 n +1

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• Fall '08
• Akhmedov,A
• #, Xn, Dominated convergence theorem, Monotone convergence theorem, David Joseph Stith

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