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vrbke1007kkr
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Homework Statement
let sn be a non-negative sequence, and define [tex]\sigma[/tex]n to be 1/n*(s1+s2+...sn)
Show that lim inf sn <= lim inf [tex]\sigma[/tex]n and lim sup[tex]\sigma[/tex]n <= lim sup sn
Homework Equations
Hint: For the last inequality show that M>N imply: sup{[tex]\sigma[/tex]n:n>M} <= 1/M*(s1+s2+...sN) + sup{sn:n>N}
The Attempt at a Solution
So one fact I could gather is that [tex]\sigma[/tex]n <= Max{s1,s2...sn} since the average can't be higher than the highest member value. Also for M>N, let sup{s_n:n>N} = sm, then the average is maximized at m, and if m<M then sup{[tex]\sigma[/tex]n:n>M} < [tex]\sigma[/tex]m, since all elements of the set of averages are forced to take elements smaller than sm such as sm+1, sm+2... sM. But am I on the right track? And can I use this information to deduce what was given in the Hint.
Also, even when I have the hint how would I connect it with lim sup? In other words, what would be the index going to infinity for the sequence 1/M*(s1+s2+...sN)?