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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
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
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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