Tuesday, 26 April 2016

calculus - Baby Rudin ("Principles of Mathematical Analysis"), chapter 3 exercise 3




If s1=2,
and sn+1=2+sn




Prove that sn is converging, and that sn<2 for all n=1,2



My attempt to use mathematical induction: if sn<2 then I use
sn+1=2+sn, to prove sn+1<2+2=2.
I guess sn is an increasing sequence, I can use the sn is bounded and increasing, and then prove that sn converges.


Answer



To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.





Lemma: sn is bounded




Base case:
n=1



s1=2<2



Suppose now for arbitrary k




sk<2



Now observe as the inductive step:
sk+1=2+sk<2+2=2



By the principle of mathematical induction we have found that 2 is a bound for all nN.




Lemma: sn is increasing





Base case:
s2s1>0
2+22>2+02=0



Now suppose for arbitrary k we have:
sksk1>0sk>sk1sk>sk1



We now make our inductive step:

sk+1sk=2+sk2+sk1>2+sk12+sk1=0
Hence we have an increasing sequence for all nN this completes the proof and our sequence is indeed convergent .


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