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 n∈N.
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+1−2=(an+1−an)2=(1an−an2)2=(a2n−22an)2>0,
so an>√2 for every n>1. If we set bn=a2n+1−2, we have b1=14,bn>0 and:
bn=b2n−14a2n=b2n−14(bn−1+2)≤b2n−18,
from which it follows that the sequence {bn}n≥1 decreases very fast towards zero. For instance, b1=14 and the last inequality imply, by induction:
bn≤125⋅2n−1−3.
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