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Lemma 14.11 Let Sn = integer n. n k=1 = 1+ 712 1 n · + −, where n € N. Then S2 ≥ 1+ n for every positive

Lemma 14.11 Let Sn = integer n. n k=1 = 1+ 712 1 n · + −, where n € N. Then S2 ≥ 1+ n for every positive

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Could you please elaborate the following proof more? I understand the basis step, and I get what we're trying to proof in the induction step, my problem is the highlighted part, things are being added and substituted and I have no idea where they come from, could you please explain in detail what they are doing in these steps, what steps they're taking? I want to understand this exact proof

Lemma 14.11 Let Sn = integer n. Proof n k=1 = 1+ 1 1 n +-, where ne N. Then S2 ≥1+ for every positive 2 n We proceed by induction. For n = 1, $₂¹ = 1 + and so the result holds for n = 1. 1 2 k , where ke N. We show that S₂k+1 ≥ 1+ Assume that s₂k ≥ 1+ that 2 $2k+1 = 1+ 1 2 = $₂k + ≥ S₂k + = S₂k + + 1 2k + 1 1 2k+1 2k 2k+1 1 2k+1 1 2k +2 1 2k+1 + + + S2k + k 1 ≥1+=+ =1+ 2 2 + 1 + 2 k + 1 2 + 1 2k+1 1 ok+1 By the Principle of Mathematical Induction, S2″ ≥1+ integer n. n IN k+1 2 Now observe for every positive
Transcribed Image Text:Lemma 14.11 Let Sn = integer n. Proof n k=1 = 1+ 1 1 n +-, where ne N. Then S2 ≥1+ for every positive 2 n We proceed by induction. For n = 1, $₂¹ = 1 + and so the result holds for n = 1. 1 2 k , where ke N. We show that S₂k+1 ≥ 1+ Assume that s₂k ≥ 1+ that 2 $2k+1 = 1+ 1 2 = $₂k + ≥ S₂k + = S₂k + + 1 2k + 1 1 2k+1 2k 2k+1 1 2k+1 1 2k +2 1 2k+1 + + + S2k + k 1 ≥1+=+ =1+ 2 2 + 1 + 2 k + 1 2 + 1 2k+1 1 ok+1 By the Principle of Mathematical Induction, S2″ ≥1+ integer n. n IN k+1 2 Now observe for every positive
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I really only need the highlighted part explained, if you could explain how they take the steps as in the highlighted part, im new to proofs.

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