transgalactic said:
i am given a siquence An [sequence?]
how to prove this EQUATION
lim sup(1/An)= 1/(lim inf An)
Let m = lim inf An. Then
1. m is a cluster point (aka limit point, aka accumulation point, etc.) of An.
That means given epsilon(e), and given any N, there exists k > N such that Ak < m + e.
2. m is the lowest such point. If m2 < m, then there exists M such that all Ak after A[M] are > m2 + e.
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REVERSE: lim sup(..) is the largest cluster point.
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Now if m,m1,m2,.. are the cluster points of An, then what are the cluster points of 1/An?
They should be 1/m, 1/m1, etc.
let's hope that none of them is zero. Then 1/m is the largest of them. And 1/m is then lim sup(1/An)
