Sunday, 11 November 2018

calculus - Show that $a_n



We define a1=1 and an+1=12(an+2an).



I want to prove that an converges. But first I'd like to show that an<2 for every nN.



I try to use induction. For a fixed n, I suppose that an<2.



Any hint to prove that an+1=12(an+2an)<2?




Thanks.


Answer



We have:
2anan+1=a2n+2,


from which it follows that:
a2n+12=(an+1an)2=(1anan2)2=(a2n22an)2>0,

so an>2 for every n>1. If we set bn=a2n+12, we have b1=14,bn>0 and:
bn=b2n14a2n=b2n14(bn1+2)b2n18,

from which it follows that the sequence {bn}n1 decreases very fast towards zero. For instance, b1=14 and the last inequality imply, by induction:
bn1252n13.



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